RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1F

 Exercise 1F

PAGE NO-43

Question 1:

Write the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$$\frac{1}{\sqrt{2}+\sqrt{3}}$.

Answer 1:

$\frac{1}{\sqrt{2}+\sqrt{3}}$$\frac{1}{\sqrt{2}+\sqrt{3}}$
$=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}$$$\begin{gathered} =\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}$$$\begin{gathered} =\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\frac{\sqrt{3}-\sqrt{2}}{1} \end{gathered}$$
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$$\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$$\sqrt{3}-\sqrt{2}$.

Question 2:

Rationalise the denominator of each of the following.
(i) $\frac{1}{\sqrt{7}}$$\frac{{\displaystyle 1}}{{\displaystyle \sqrt{7}}}$                     (ii) $\frac{\sqrt{5}}{2\sqrt{3}}$$\frac{{\displaystyle \sqrt{5}}}{{\displaystyle 2\sqrt{3}}}$                   (iii) $\frac{1}{2+\sqrt{3}}$$\frac{1}{2+\sqrt{3}}$
(iv) $\frac{1}{\sqrt{5}-2}$$\frac{1}{\sqrt{5}{\displaystyle -}{\displaystyle 2}}$               (v) $\frac{1}{5+3\sqrt{2}}$$\frac{1}{5+3\sqrt{2}}$              (vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$$\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}$
(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$$\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}$        (viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$$\frac{1+\sqrt{2}}{2-\sqrt{2}}$              (ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$

Answer 2:

(i) $\frac{1}{\sqrt{7}}$$\frac{{\displaystyle 1}}{{\displaystyle \sqrt{7}}}$
On multiplying the numerator and denominator of the given number by $\sqrt{7}$$\sqrt{7}$, we get:

 $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$$\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

(ii) $\frac{\sqrt{5}}{2\sqrt{3}}$$\frac{{\displaystyle \sqrt{5}}}{{\displaystyle 2\sqrt{3}}}$
On multiplying the numerator and denominator of the given number by $\sqrt{3}$$\sqrt{3}$, we get:

 $\frac{\sqrt{5}}{2\sqrt{3}}=\frac{\sqrt{5}}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{6}$$\frac{\sqrt{5}}{2\sqrt{3}} = \frac{\sqrt{5}}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{15}}{6}$

(iii) $\frac{1}{2+\sqrt{3}}$$\frac{1}{2+\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$$2-\sqrt{3}$, we get:
 $\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$$$\begin{gathered} \frac{1}{2+\sqrt{3}} = \frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ =\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ = \frac{2-\sqrt{3}}{4-3} \\ =\frac{2-\sqrt{3}}{1} \\ = 2-\sqrt{3} \end{gathered}$$

(iv) $\frac{1}{\sqrt{5}-2}$$\frac{1}{\sqrt{5}{\displaystyle -}{\displaystyle 2}}$
On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$$\sqrt{5}+2$, we get:
 $\frac{1}{\sqrt{5}-2}=\frac{1}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2}=\frac{\sqrt{5}+2}{5-4}=\frac{\sqrt{5}+2}{1}=\sqrt{5}+2$$$\begin{gathered} \frac{1}{\sqrt{5}-2} = \frac{1}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2} \\ =\frac{\sqrt{5}+2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2} \\ = \frac{\sqrt{5}+2}{5-4} \\ =\frac{\sqrt{5}+2}{1} \\ = \sqrt{5}+2 \end{gathered}$$

(v) $\frac{1}{5+3\sqrt{2}}$$\frac{1}{5+3\sqrt{2}}$
On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$$5-3\sqrt{2}$, we get:
 $\frac{1}{5+3\sqrt{2}}=\frac{1}{5+3\sqrt{2}}\times \frac{5-3\sqrt{2}}{5-3\sqrt{2}}=\frac{5-3\sqrt{2}}{{\left(5\right)}^2-{\left(3\sqrt{2}\right)}^2}=\frac{5-3\sqrt{2}}{25-18}=\frac{5-3\sqrt{2}}{7}$$$\begin{gathered} \frac{1}{5+3\sqrt{2}} = \frac{1}{5+3\sqrt{2}}\times \frac{5-3\sqrt{2}}{5-3\sqrt{2}} \\ =\frac{5-3\sqrt{2}}{{\left(5\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ = \frac{5-3\sqrt{2}}{25-18} \\ =\frac{5-3\sqrt{2}}{7} \end{gathered}$$

(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$$\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}$
Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$$\sqrt{7}+\sqrt{6}$, we get
$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{6}\right)}^2}$$$\begin{gathered} \frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}=\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}{\displaystyle +}{\displaystyle \sqrt{6}}} \\ =\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^2{\displaystyle -{\left(\sqrt{6}\right)}^2}} \end{gathered}$$
$=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}$$$\begin{gathered} =\frac{\sqrt{7}+\sqrt{6}}{{\displaystyle 7-6}} \\ =\sqrt{7}+\sqrt{6} \end{gathered}$$

(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$$\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}$     
Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$$\sqrt{11}+\sqrt{7}$, we get
$\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}}\times \frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^2-{\left(\sqrt{7}\right)}^2}$$$\begin{gathered} \frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}=\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}\times \frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}{\displaystyle +}{\displaystyle \sqrt{7}}} \\ =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{7}\right)}^2}} \end{gathered}$$ 
$=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11-7}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4}=\sqrt{11}+\sqrt{7}$$$\begin{gathered} =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11{\displaystyle -}{\displaystyle 7}} \\ =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4} \\ =\sqrt{11}+\sqrt{7} \end{gathered}$$

(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$$\frac{1+\sqrt{2}}{2-\sqrt{2}}$ 
Multiplying the numerator and denominator by $2+\sqrt{2}$$2+\sqrt{2}$, we get 
$\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}$$$\begin{gathered} \frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ =\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$        
$=\frac{4+3\sqrt{2}}{4-2}=\frac{4+3\sqrt{2}}{2}$$$\begin{gathered} =\frac{4+3\sqrt{2}}{4-2} \\ =\frac{4+3\sqrt{2}}{2} \end{gathered}$$
(ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Multiplying the numerator and denominator by $3-2\sqrt{2}$$3-2\sqrt{2}$, we get 
$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}=\frac{{\left(3-2\sqrt{2}\right)}^2}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}$$$\begin{gathered} \frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ =\frac{{\left(3-2\sqrt{2}\right)}^2}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{9+8-12\sqrt{2}}{9-8} \left[{\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ =17-12\sqrt{2} \end{gathered}$$

Question 3:

It being given that $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236 \mathrm{and} \sqrt{10}=3.162$, find the value of three places of decimals, of each of the following.
(i) $\frac{2}{\sqrt{5}}$

(ii) $\frac{{\displaystyle 2-\sqrt{3}}}{{\displaystyle \sqrt{3}}}$

(iii) $\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$

Answer 3:

(i)
 $$\begin{gathered} \frac{2}{\sqrt{5}} \\ =\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ =\frac{2\sqrt{5}}{5} \end{gathered}$$
$$\begin{gathered} =\frac{2\times 2.236}{5} \\ =0.894 \end{gathered}$$
(ii) 
$$\begin{gathered} \frac{{\displaystyle 2-\sqrt{3}}}{{\displaystyle \sqrt{3}}} \\ =\frac{{\displaystyle 2-\sqrt{3}}}{{\displaystyle \sqrt{3}}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{2\sqrt{3}-3}{3} \end{gathered}$$
$$\begin{gathered} =\frac{2\times 1.732-3}{3} \\ =0.155 \end{gathered}$$
(iii)
 $$\begin{gathered} \frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}} \\ =\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\sqrt{2}\left(\sqrt{10}-\sqrt{5}\right)}{2} \end{gathered}$$
$$\begin{gathered} =\frac{1.414\times \left(3.162-2.236\right)}{2} \\ =0.655 \end{gathered}$$

Question 4:

Find rational numbers a and b such that

(i) $\frac{\sqrt{2}-1}{\sqrt{2}+1}=a+b\sqrt{2}$

(ii) $\frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}=a\sqrt{5}+b$

(iii) $\frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}=a+b\sqrt{6}$

(iv) $\frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}=a+b\sqrt{3}$

Answer 4:

(i)
$$\begin{gathered} \frac{\sqrt{2}-1}{\sqrt{2}+1} \\ =\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1} \\ =\frac{{\left(\sqrt{2}-1\right)}^2}{{\left(\sqrt{2}\right)}^2-1^2} \end{gathered}$$
$$\begin{gathered} =\frac{2+1-2\sqrt{2}}{2-1} \\ =3-2\sqrt{2} \end{gathered}$$
$$\begin{gathered} \therefore \frac{\sqrt{2}-1}{\sqrt{2}+1}=3+\left(-2\right)\sqrt{2}=a+b\sqrt{2} \\ \Rightarrow a=3, b=-2 \end{gathered}$$

(ii)
$$\begin{gathered} \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}} \\ =\frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}\times \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2-\sqrt{5}}} \\ =\frac{{\left(2-\sqrt{5}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{4+5-4\sqrt{5}}{4-5} \\ =\frac{9-4\sqrt{5}}{-1} \\ =-9+4\sqrt{5} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}=4\sqrt{5}+\left(-9\right)=a\sqrt{5}+b \\ \Rightarrow a=4, b=-9 \end{gathered}$$

(iii)
$$\begin{gathered} \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}} \\ =\frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}\times \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}+\sqrt{2}}} \\ =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{3+2+2\times \sqrt{3}\times \sqrt{2}}{3-2} \\ =5+2\sqrt{6} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}=5+2\sqrt{6}=a+b\sqrt{6} \\ \Rightarrow a=5, b=2 \end{gathered}$$

(iv)
$$\begin{gathered} \frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}} \\ =\frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}\times \frac{{\displaystyle 7-4\sqrt{3}}}{{\displaystyle 7-4\sqrt{3}}} \\ =\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{11-6\sqrt{3}}{49-48} \\ =11-6\sqrt{3} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3} \\ \Rightarrow a=11,b=-6 \end{gathered}$$

Question 5:

It being given  that $\sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{6}=2.449 \mathrm{and} \sqrt{10}=3.162$, find to three places of decimal, the value of each of the following.

(i) $\frac{1}{\sqrt{6}+\sqrt{5}}$

(ii) $\frac{{\displaystyle 6}}{{\displaystyle \sqrt{5}+\sqrt{3}}}$

(iii) $\frac{{\displaystyle 1}}{{\displaystyle 4\sqrt{3}-3\sqrt{5}}}$

(iv) $\frac{3+\sqrt{5}}{3-\sqrt{5}}$

(v) $\frac{1+2\sqrt{3}}{2-\sqrt{3}}$

(vi) $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

Answer 5:

(i)
 $$\begin{gathered} \frac{1}{\sqrt{6}+\sqrt{5}} \\ =\frac{1}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\ =\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ =\sqrt{6}-\sqrt{5} \\ =2.449-2.236 \end{gathered}$$
$=0.213$
(ii) 
$$\begin{gathered} \frac{{\displaystyle 6}}{{\displaystyle \sqrt{5}+\sqrt{3}}} \\ =\frac{{\displaystyle 6}}{{\displaystyle \sqrt{5}+\sqrt{3}}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \\ =\frac{{\displaystyle 6\left(\sqrt{5}-\sqrt{3}\right)}}{{\displaystyle {\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{5-3} \\ =\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{2} \\ =3\left(\sqrt{5}-\sqrt{3}\right) \end{gathered}$$
$$\begin{gathered} =3\times \left(2.236-1.732\right) \\ =1.512 \end{gathered}$$
(iii)
 $$\begin{gathered} \frac{{\displaystyle 1}}{{\displaystyle 4\sqrt{3}-3\sqrt{5}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 4\sqrt{3}-3\sqrt{5}}}\times \frac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}} \\ =\frac{4\sqrt{3}+3\sqrt{5}}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{4\sqrt{3}+3\sqrt{5}}{48-45} \\ =\frac{4\times 1.732+3\times 2.236}{3} \\ =4.545 \end{gathered}$$
(iv)
 $$\begin{gathered} \frac{3+\sqrt{5}}{3-\sqrt{5}} \\ =\frac{3+\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ =\frac{{\left(3+\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{9+5+6\sqrt{5}}{9-5} \\ =\frac{14+6\sqrt{5}}{4} \\ =\frac{7+3\sqrt{5}}{2} \end{gathered}$$
$$\begin{gathered} =\frac{7+3\times 2.236}{2} \\ =6.854 \end{gathered}$$
(v)
 $$\begin{gathered} \frac{1+2\sqrt{3}}{2-\sqrt{3}} \\ =\frac{1+2\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\ =\frac{2+\sqrt{3}+4\sqrt{3}+6}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{8+5\sqrt{3}}{4-3} \\ =8+5\sqrt{3} \\ =8+5\times 1.732 \end{gathered}$$
$=16.660$
(vi)
 $$\begin{gathered} \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{5+2+2\times \sqrt{5}\times \sqrt{2}}{5-2} \\ =\frac{7+2\sqrt{10}}{3} \\ =\frac{7+2\times 3.162}{3} \end{gathered}$$PAGE NO -44$=4.441$

Question 6:

Simplify by rationalising the denominator.

(i) $\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

(ii) $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$

Answer 6:

(i) 
$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16\times 3}+\sqrt{9\times 2}}=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$$$\begin{gathered} \frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}} \\ =\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16\times 3}+\sqrt{9\times 2}} \\ =\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}} \end{gathered}$$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}=\frac{7\sqrt{3}\times 4\sqrt{3}-7\sqrt{3}\times 3\sqrt{2}-5\sqrt{2}\times 4\sqrt{3}+5\sqrt{2}\times 3\sqrt{2}}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2}=\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-18}$$$\begin{gathered} =\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}} \\ =\frac{7\sqrt{3}\times 4\sqrt{3}-7\sqrt{3}\times 3\sqrt{2}-5\sqrt{2}\times 4\sqrt{3}+5\sqrt{2}\times 3\sqrt{2}}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-18} \end{gathered}$$
$=\frac{114-41\sqrt{6}}{30}$$=\frac{114-41\sqrt{6}}{30}$
(ii)
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times \frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}=\frac{2\sqrt{6}\times 3\sqrt{5}+2\sqrt{6}\times 2\sqrt{6}-\sqrt{5}\times 3\sqrt{5}-\sqrt{5}\times 2\sqrt{6}}{{\left(3\sqrt{5}\right)}^2-{\left(2\sqrt{6}\right)}^2}$$$\begin{gathered} \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}} \\ =\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times \frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}} \\ =\frac{2\sqrt{6}\times 3\sqrt{5}+2\sqrt{6}\times 2\sqrt{6}-\sqrt{5}\times 3\sqrt{5}-\sqrt{5}\times 2\sqrt{6}}{{\left(3\sqrt{5}\right)}^2-{\left(2\sqrt{6}\right)}^2} \end{gathered}$$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}=\frac{9+4\sqrt{30}}{21}$$$\begin{gathered} =\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24} \\ =\frac{9+4\sqrt{30}}{21} \end{gathered}$$

Question 7:

Simplify

(i) $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$

(ii) $\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}$$\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{2}}}-\frac{2}{\sqrt{5}{\displaystyle -}{\displaystyle \sqrt{3}}}-\frac{3}{\sqrt{2}{\displaystyle -}{\displaystyle \sqrt{5}}}$

(iii) $\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}{\displaystyle -}{\displaystyle 1}}{\sqrt{3}{\displaystyle +}{\displaystyle 1}}$

(iv) $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}$$\frac{2\sqrt{6}}{\sqrt{2}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{6\sqrt{2}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{3}}}-\frac{8\sqrt{3}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{2}}}$

Answer 7:

(i)
$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}\times \frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\times \frac{4-\sqrt{5}}{4-\sqrt{5}}=\frac{{\left(4+\sqrt{5}\right)}^2}{{\left(4\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{{\left(4-\sqrt{5}\right)}^2}{{\left(4\right)}^2-{\left(\sqrt{5}\right)}^2}$$$\begin{gathered} \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}} \\ =\frac{4+\sqrt{5}}{4-\sqrt{5}}\times \frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\times \frac{4-\sqrt{5}}{4-\sqrt{5}} \\ =\frac{{\left(4+\sqrt{5}\right)}^2}{{\left(4\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{{\left(4-\sqrt{5}\right)}^2}{{\left(4\right)}^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$=\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5}=\frac{42}{11}$$$\begin{gathered} =\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5} \\ =\frac{42}{11} \end{gathered}$$
(ii)
$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{5}\right)}^2}$$$\begin{gathered} \frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{2}}}-\frac{2}{\sqrt{5}{\displaystyle -}{\displaystyle \sqrt{3}}}-\frac{3}{\sqrt{2}{\displaystyle -}{\displaystyle \sqrt{5}}} \\ =\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{2}}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}{\displaystyle -}{\displaystyle \sqrt{3}}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}{\displaystyle -}{\displaystyle \sqrt{5}}}\times \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}} \\ =\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{3}\right)}^2}}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{{\left(\sqrt{2}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{5}\right)}^2}} \end{gathered}$$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{2-5}=\sqrt{3}-\sqrt{2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{\left(-3\right)}=\sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5}$$$\begin{gathered} =\frac{\sqrt{3}{\displaystyle -}{\displaystyle \sqrt{2}}}{3-2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{2-5} \\ =\sqrt{3}-\sqrt{2}-\frac{{\displaystyle 2\left(\sqrt{5}+\sqrt{3}\right)}}{{\displaystyle 2}}-\frac{{\displaystyle 3\left(\sqrt{2}+\sqrt{5}\right)}}{{\displaystyle \left(-3\right)}} \\ =\sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5} \end{gathered}$$
$=0$$=0$
(iii)
$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{{\left(2+\sqrt{3}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{{\left(2-\sqrt{3}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-1^2}$$$\begin{gathered} \frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}{\displaystyle -}{\displaystyle 1}}{\sqrt{3}{\displaystyle +}{\displaystyle 1}} \\ =\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}{\displaystyle -}{\displaystyle 1}}{\sqrt{3}{\displaystyle +}{\displaystyle 1}}\times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\ =\frac{{\left(2+\sqrt{3}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{{\left(2-\sqrt{3}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{{\left(\sqrt{3}{\displaystyle -}{\displaystyle 1}\right)}^2}{{\left(\sqrt{3}\right)}^2-1^2} \end{gathered}$$
$=\frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1}=7+4\sqrt{3}+7-4\sqrt{3}+\frac{4-2\sqrt{3}}{2}=14+2-\sqrt{3}$$$\begin{gathered} =\frac{4+3+4\sqrt{3}}{4-3}+\frac{{\displaystyle 4+3-4\sqrt{3}}}{{\displaystyle 4-3}}+\frac{3+1-2\sqrt{3}}{3-1} \\ =7+4\sqrt{3}+7-4\sqrt{3}+\frac{{\displaystyle 4-2\sqrt{3}}}{{\displaystyle 2}} \\ =14+2-\sqrt{3} \end{gathered}$$
$=16-\sqrt{3}$$=16-\sqrt{3}$
(iv)
$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}\times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{6}\times \sqrt{3}-2\sqrt{6}\times \sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}+\frac{6\sqrt{2}\times \sqrt{6}-6\sqrt{2}\times \sqrt{3}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{3}\right)}^2}-\frac{8\sqrt{3}\times \sqrt{6}-8\sqrt{3}\times \sqrt{2}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2}$$$\begin{gathered} \frac{2\sqrt{6}}{\sqrt{2}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{6\sqrt{2}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{3}}}-\frac{8\sqrt{3}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{2}}} \\ =\frac{2\sqrt{6}}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{2}}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{3}}}\times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{2}}}\times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} \\ =\frac{2\sqrt{6}\times \sqrt{3}-2\sqrt{6}\times \sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}+\frac{6\sqrt{2}\times \sqrt{6}-6\sqrt{2}\times \sqrt{3}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{3}\right)}^2}-\frac{8\sqrt{3}\times \sqrt{6}-8\sqrt{3}\times \sqrt{2}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$=\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2}=2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4}=2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6}$$$\begin{gathered} =\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2} \\ =2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4} \\ =2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6} \end{gathered}$$
$=0$$=0$

Question 8:

Prove that
(i) $\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1$$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}{\displaystyle +}{\displaystyle \sqrt{5}}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle 1}}=1$
(ii) $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}=2$$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{4}}}+\frac{1}{\sqrt{4}{\displaystyle +}{\displaystyle \sqrt{5}}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{6}}}+\frac{1}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{7}}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}=2$

Answer 8:

(i)
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=\frac{1}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3-\sqrt{7}}{{\left(3\right)}^2-{\left(\sqrt{7}\right)}^2}+\frac{\sqrt{7}-\sqrt{5}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{\sqrt{3}-1}{{\left(\sqrt{3}\right)}^2-1^2}$$$\begin{gathered} \frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}{\displaystyle +}{\displaystyle \sqrt{5}}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle 1}} \\ =\frac{1}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}{\displaystyle +}{\displaystyle \sqrt{5}}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{3}}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle 1}}\times \frac{\sqrt{3}-1}{\sqrt{3}-{\displaystyle 1}} \\ =\frac{3-\sqrt{7}}{{\left(3\right)}^2-{\left(\sqrt{7}\right)}^2}+\frac{\sqrt{7}-\sqrt{5}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{\sqrt{3}-1}{{\left(\sqrt{3}\right)}^2-{\displaystyle 1^2}} \end{gathered}$$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$$$\begin{gathered} =\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-{\displaystyle 1}} \\ =\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2} \\ =\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2} \end{gathered}$$
$=\frac{2}{2}=1$$$\begin{gathered} =\frac{2}{2} \\ =1 \end{gathered}$$
(ii)
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}=\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\times \frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\times \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}\times \frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}$$$\begin{gathered} \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}{\displaystyle +}{\displaystyle \sqrt{3}}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{4}}}+\frac{1}{\sqrt{4}{\displaystyle +}{\displaystyle \sqrt{5}}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{6}}}+\frac{1}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{7}}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}} \\ =\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}{\displaystyle +}{\displaystyle \sqrt{3}}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-{\displaystyle \sqrt{3}}}+\frac{1}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{4}}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}{\displaystyle +}{\displaystyle \sqrt{5}}}\times \frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}{\displaystyle +}{\displaystyle \sqrt{6}}}\times \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}{\displaystyle +}{\displaystyle \sqrt{7}}}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}\times \frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}} \end{gathered}$$
$=\frac{1-\sqrt{2}}{1^2-{\left(\sqrt{2}\right)}^2}+\frac{\sqrt{2}-\sqrt{3}}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{\sqrt{3}-\sqrt{4}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{4}\right)}^2}+\frac{\sqrt{4}-\sqrt{5}}{{\left(\sqrt{4}\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{\sqrt{5}-\sqrt{6}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{6}\right)}^2}+\frac{\sqrt{6}-\sqrt{7}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{7}\right)}^2}+\frac{\sqrt{7}-\sqrt{8}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{8}\right)}^2}+\frac{\sqrt{8}-\sqrt{9}}{{\left(\sqrt{8}\right)}^2-{\left(\sqrt{9}\right)}^2}=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}=\frac{1-\sqrt{2}}{\left(}+\frac{\sqrt{2}-\sqrt{3}}{\left(}+\frac{\sqrt{3}-\sqrt{4}}{\left(}+\frac{\sqrt{4}-\sqrt{5}}{\left(}+\frac{\sqrt{5}-\sqrt{6}}{\left(}+\frac{\sqrt{6}-\sqrt{7}}{\left(}+\frac{\sqrt{7}-\sqrt{8}}{\left(}+\frac{\sqrt{8}-\sqrt{9}}{\left(}$$$\begin{gathered} =\frac{1-\sqrt{2}}{1^2-{\left(\sqrt{2}\right)}^2}+\frac{\sqrt{2}-\sqrt{3}}{{\left(\sqrt{2}\right)}^2-{\displaystyle {\left(\sqrt{3}\right)}^2}}+\frac{\sqrt{3}-\sqrt{4}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{4}\right)}^2}+\frac{\sqrt{4}-\sqrt{5}}{{\left(\sqrt{4}\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{\sqrt{5}-\sqrt{6}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{6}\right)}^2}+\frac{\sqrt{6}-\sqrt{7}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{7}\right)}^2}+\frac{\sqrt{7}-\sqrt{8}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{8}\right)}^2}+\frac{\sqrt{8}-\sqrt{9}}{{\left(\sqrt{8}\right)}^2-{\left(\sqrt{9}\right)}^2} \\ =\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-{\displaystyle 3}}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9} \\ =\frac{1-\sqrt{2}}{\left(-1\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(-1\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(-1\right)}+\frac{\sqrt{4}-\sqrt{5}}{\left(-1\right)}+\frac{\sqrt{5}-\sqrt{6}}{\left(-1\right)}+\frac{\sqrt{6}-\sqrt{7}}{\left(-1\right)}+\frac{\sqrt{7}-\sqrt{8}}{\left(-1\right)}+\frac{\sqrt{8}-\sqrt{9}}{\left(-1\right)} \end{gathered}$$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}=3-1=2$$$\begin{gathered} =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8} \\ =3-1 \\ =2 \end{gathered}$$

Question 9:

Find the values of a and b if
$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+b\sqrt{5}$

Answer 9:

7+3√53+√5-7-3√53-√5=7+3√53+√5×3-√53-√5-7-3√53-√5×3+√53+√5=7(3-√5)+3√5(3-√5)32-(√5)2-7(3+√5)-3√5(3+√5)32-(√5)2$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2}$$$\begin{gathered} \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}} \\ =\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ =\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
=21-7√5+9√5-159-5-21+7√5-9√5-159-5=6+2√54-6-2√54$=\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5}=\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4}$$$\begin{gathered} =\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5} \\ =\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4} \end{gathered}$$
=6+2√5-6+2√54=4√54=√5$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}=\frac{4\sqrt{5}}{4}=\sqrt{5}$$$\begin{gathered} =\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4} \\ =\frac{4\sqrt{5}}{4} \\ =\sqrt{5} \end{gathered}$$
∴7+3√53+√5-7-3√53-√5=0+1×√5$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1\times \sqrt{5}$$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1\times \sqrt{5}$
Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.

Question 10:

Simplify $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}{\displaystyle +}{\displaystyle \sqrt{11}}}{\sqrt{13}{\displaystyle -}{\displaystyle \sqrt{11}}}$.

Answer 10:

$$\begin{gathered} \frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}{\displaystyle +}{\displaystyle \sqrt{11}}}{\sqrt{13}{\displaystyle -}{\displaystyle \sqrt{11}}} \\ =\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}\times \frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}+\frac{\sqrt{13}{\displaystyle +}{\displaystyle \sqrt{11}}}{\sqrt{13}{\displaystyle -}{\displaystyle \sqrt{11}}}\times \frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}} \\ =\frac{{\left(\sqrt{13}-\sqrt{11}\right)}^2}{{\left(\sqrt{13}\right)}^2-{\left(\sqrt{11}\right)}^2}+\frac{{\left(\sqrt{13}+\sqrt{11}\right)}^2}{{\left(\sqrt{13}\right)}^2-{\left(\sqrt{11}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{13+11-2\times \sqrt{13}\times \sqrt{11}}{13-11}+\frac{13+11+2\times \sqrt{13}\times \sqrt{11}}{13-11} \\ =\frac{24-2\sqrt{143}}{2}+\frac{24+2\sqrt{143}}{2} \\ =\frac{24-2\sqrt{143}+24+2\sqrt{143}}{2} \end{gathered}$$
$$\begin{gathered} =\frac{48}{2} \\ =24 \end{gathered}$$

Question 11:

If $x=3+2\sqrt{2}$, check whether $x+\frac{1}{x}$ is rational or irrational.

Answer 11:

$x=3+2\sqrt{2} .....\left(1\right)$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{1}{3+2\sqrt{2}} \\ \Rightarrow \frac{1}{x}=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ \Rightarrow \frac{1}{x}=\frac{3-2\sqrt{2}}{3^2-{\left(2\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{3-2\sqrt{2}}{9-8} \\ \Rightarrow \frac{1}{x}=3-2\sqrt{2} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$, which is a rational number
Thus, $x+\frac{1}{x}$ is rational.

Question 12:

If $x=2-\sqrt{3}$, find value of ${\left(x-\frac{1}{x}\right)}^3$.

Answer 12:

$$\begin{gathered} x=2-\sqrt{3} .....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2+\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2+\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{x}=2+\sqrt{3} .....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} x-\frac{1}{x}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right) \\ \Rightarrow x-\frac{1}{x}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3} \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^3={\left(-2\sqrt{3}\right)}^3=-24\sqrt{3} \end{gathered}$$
Thus, the value of ${\left(x-\frac{1}{x}\right)}^3$ is $-24\sqrt{3}$.

Question 13:

If $x=9-4\sqrt{5}$, find the value of $x^2+\frac{1}{x^2}$.

Answer 13:

$$\begin{gathered} x=9-4\sqrt{5} .....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{9-4\sqrt{5}} \\ \Rightarrow \frac{1}{x}=\frac{1}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{9+4\sqrt{5}}{9^2-{\left(4\sqrt{5}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{9+4\sqrt{5}}{81-80} \\ \Rightarrow \frac{1}{x}=9+4\sqrt{5} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5} \\ \Rightarrow x+\frac{1}{x}=18 \end{gathered}$$
Squaring on both sides, we get
$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={18}^2 \\ \Rightarrow x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=324 \\ \Rightarrow x^2+\frac{1}{x^2}=324-2=322 \end{gathered}$$
Thus, the value of $x^2+\frac{1}{x^2}$ is 322.

Question 14:

If $x=\frac{5-\sqrt{21}}{2}$, find the value of $x+\frac{1}{x}$.

Answer 14:

$$\begin{gathered} x=\frac{5-\sqrt{21}}{2} .....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{{\displaystyle \frac{5-\sqrt{21}}{2}}} \\ \Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}}\times \frac{5+\sqrt{21}}{5+\sqrt{21}} \\ \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{5^2-{\left(\sqrt{21}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{25-21} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{4} \\ \Rightarrow \frac{1}{x}=\frac{5+\sqrt{21}}{2} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2} \\ \Rightarrow x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2} \\ \Rightarrow x+\frac{1}{x}=\frac{10}{2}=5 \end{gathered}$$
Thus, the value of $x+\frac{1}{x}$ is 5.

Question 15:

If $a=3-2\sqrt{2}$, find the value of $a^2-\frac{1}{a^2}$.

Answer 15:

$$\begin{gathered} a=3-2\sqrt{2} \\ \Rightarrow a^2={\left(3-2\sqrt{2}\right)}^2 \\ \Rightarrow a^2=9+8-12\sqrt{2} \\ \Rightarrow a^2=17-12\sqrt{2} .....\left(1\right) \end{gathered}$$
$$\begin{gathered} \therefore \frac{1}{a^2}=\frac{1}{17-12\sqrt{2}} \\ \Rightarrow \frac{1}{a^2}=\frac{1}{17-12\sqrt{2}}\times \frac{17+12\sqrt{2}}{17+12\sqrt{2}} \\ \Rightarrow \frac{1}{a^2}=\frac{17+12\sqrt{2}}{{17}^2-{\left(12\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{a^2}=\frac{17+12\sqrt{2}}{289-288} \\ \Rightarrow \frac{1}{a^2}=17+12\sqrt{2} .....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} a^2-\frac{1}{a^2}=\left(17-12\sqrt{2}\right)-\left(17+12\sqrt{2}\right) \\ \Rightarrow a^2-\frac{1}{a^2}=17-12\sqrt{2}-17-12\sqrt{2} \\ \Rightarrow a^2-\frac{1}{a^2}=-24\sqrt{2} \end{gathered}$$
Thus, the value of $a^2-\frac{1}{a^2}$ is $-24\sqrt{2}$.

PAGE NO-45

Question 16:

If $x=\sqrt{13}+2\sqrt{3}$, find the value of $x-\frac{1}{x}$.

Answer 16:

$$\begin{gathered} x=\sqrt{13}+2\sqrt{3} .....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2\sqrt{3}}\times \frac{\sqrt{13}-2\sqrt{3}}{\sqrt{13}-2\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2\sqrt{3}}{{\left(\sqrt{13}\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2\sqrt{3}}{13-12} \\ \Rightarrow \frac{1}{x}=\sqrt{13}-2\sqrt{3} .....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} x-\frac{1}{x}=\left(\sqrt{13}+2\sqrt{3}\right) -\left(\sqrt{13}-2\sqrt{3} \right) \\ \Rightarrow x-\frac{1}{x}=\sqrt{13}+2\sqrt{3} -\sqrt{13}+2\sqrt{3} \\ \Rightarrow x-\frac{1}{x}=4\sqrt{3} \end{gathered}$$
Thus, the value of $x-\frac{1}{x}$ is $4\sqrt{3}$.

Question 17:

If $x=2+\sqrt{3}$, find the value of $x^3+\frac{1}{x^3}$.

Answer 17:

$$\begin{gathered} x=2+\sqrt{3} .....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{x}=2-\sqrt{3} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4 .....\left(3\right)$
Cubing both sides, we get
$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^3=4^3 \\ \Rightarrow x^3+\frac{1}{x^3}+3\times x\times \frac{1}{x}\left(x+\frac{1}{x}\right)=64 \end{gathered}$$
$\Rightarrow x^3+\frac{1}{x^3}+3\times 4=64$             [Using (3)]
$\Rightarrow x^3+\frac{1}{x^3}=64-12=52$
Thus, the value of $x^3+\frac{1}{x^3}$ is 52.

Question 18:

If $x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \mathrm{and} y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$, show that $x^2-y^2=-\frac{10\sqrt{3}}{11}$.

Answer 18:

Disclaimer: The question is incorrect.

$$\begin{gathered} x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \\ \Rightarrow x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\times \frac{5-\sqrt{3}}{5-\sqrt{3}} \\ \Rightarrow x=\frac{{\left(5-\sqrt{3}\right)}^2}{5^2-{\left(\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{25+3-10\sqrt{3}}{25-3} \\ \Rightarrow x=\frac{28-10\sqrt{3}}{22} \\ \Rightarrow x=\frac{14-5\sqrt{3}}{11} \end{gathered}$$

$$\begin{gathered} y=\frac{5+\sqrt{3}}{5-\sqrt{3}} \\ \Rightarrow y=\frac{5+\sqrt{3}}{5-\sqrt{3}}\times \frac{5+\sqrt{3}}{5+\sqrt{3}} \\ \Rightarrow y=\frac{{\left(5+\sqrt{3}\right)}^2}{5^2-{\left(\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow y=\frac{25+3+10\sqrt{3}}{25-3} \\ \Rightarrow y=\frac{28+10\sqrt{3}}{22} \\ \Rightarrow y=\frac{14+5\sqrt{3}}{11} \end{gathered}$$

$$\begin{gathered} \therefore x^2-y^2 \\ ={\left(\frac{14-5\sqrt{3}}{11}\right)}^2-{\left(\frac{14+5\sqrt{3}}{11}\right)}^2 \\ =\frac{196+75-140\sqrt{3}}{121}-\frac{196+75+140\sqrt{3}}{121} \end{gathered}$$
$$\begin{gathered} =\frac{271-140\sqrt{3}}{121}-\frac{271+140\sqrt{3}}{121} \\ =\frac{271-140\sqrt{3}-271-140\sqrt{3}}{121} \\ =\frac{-280\sqrt{3}}{121} \end{gathered}$$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10\sqrt{3}}{11}$.
$$\begin{gathered} \therefore x-y \\ =\frac{14-5\sqrt{3}}{11}-\frac{14+5\sqrt{3}}{11} \\ =\frac{14-5\sqrt{3}-14-5\sqrt{3}}{11} \\ =\frac{-10\sqrt{3}}{11} \end{gathered}$$

Question 19:

If $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \mathrm{and} b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, show that $3a^2+4ab-3b^2=4+\frac{56}{3}\sqrt{10}$.

Answer 19:

According to question,
$a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \mathrm{and} b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$$\begin{gathered} a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2+2\sqrt{10}}{3} \\ =\frac{7+2\sqrt{10}}{3} ...\left(1\right) \\ b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{{\left(\sqrt{5}-\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2-2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2-2\sqrt{10}}{3} \\ =\frac{7-2\sqrt{10}}{3} ...\left(2\right) \end{gathered}$$
Now,
$$\begin{gathered} 3a^2+4ab-3b^2 \\ =3\left(a^2-b^2\right)+4ab \\ =3\left(a+b\right)\left(a-b\right)+4ab \\ =3\left(\frac{7+2\sqrt{10}}{3}+\frac{7-2\sqrt{10}}{3}\right)\left(\frac{7+2\sqrt{10}}{3}-\frac{7-2\sqrt{10}}{3}\right)+4\left(\frac{7+2\sqrt{10}}{3}\times \frac{7-2\sqrt{10}}{3}\right) \\ =3\left(\frac{14}{3}\right)\left(\frac{4\sqrt{10}}{3}\right)+4\left(\frac{{\left(7\right)}^2-{\left(2\sqrt{10}\right)}^2}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4\left(\frac{49-40}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4 \end{gathered}$$
Hence, $3a^2+4ab-3b^2=4+\frac{56\sqrt{10}}{3}$.

Question 20:

If $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \mathrm{and} b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, find the value of a² + b² – 5ab.

Answer 20:

According to question,
$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \mathrm{and} b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2-2\sqrt{6}}{3-2} \\ =\frac{5-2\sqrt{6}}{1} \\ =5-2\sqrt{6} ....\left(1\right) \\ b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\left(\sqrt{3}\right)\left(\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2+2\sqrt{6}}{3-2} \\ =\frac{5+2\sqrt{6}}{1} \\ =5+2\sqrt{6} ....\left(2\right) \end{gathered}$$
Now,
$$\begin{gathered} a^2+b^2-5ab \\ ={\left(a-b\right)}^2-3ab \\ ={\left\{\left(5-2\sqrt{6}\right)-\left(5+2\sqrt{6}\right)\right\}}^2-3\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right) \\ ={\left(-4\sqrt{6}\right)}^2-3\left(25-24\right) \\ =96-3 \\ =93 \end{gathered}$$

Hence, the value of a² + b² – 5ab is 93.

Question 21:

If $p=\frac{3-\sqrt{5}}{3+\sqrt{5}} \mathrm{and} q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$, find the value of p² + q².

Answer 21:

According to question,
$p=\frac{3-\sqrt{5}}{3+\sqrt{5}} \mathrm{and} q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$$\begin{gathered} p=\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ =\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}} \\ =\frac{{\left(3-\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{{\left(3\right)}^2+{\left(\sqrt{5}\right)}^2-2\left(3\right)\left(\sqrt{5}\right)}{9-5} \\ =\frac{9+5-6\sqrt{5}}{4} \\ =\frac{14-6\sqrt{5}}{4} ...\left(1\right) \\ q=\frac{3+\sqrt{5}}{3-\sqrt{5}} \\ =\frac{3+\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ =\frac{{\left(3+\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{{\left(3\right)}^2+{\left(\sqrt{5}\right)}^2+2\left(3\right)\left(\sqrt{5}\right)}{9-5} \\ =\frac{9+5+6\sqrt{5}}{4} \\ =\frac{14+6\sqrt{5}}{4} ...\left(2\right) \end{gathered}$$

Now,
$$\begin{gathered} p^2+q^2={\left(p+q\right)}^2-2pq \\ ={\left(\frac{14-6\sqrt{5}}{4}+\frac{14+6\sqrt{5}}{4}\right)}^2-2\left(\frac{14-6\sqrt{5}}{4}\right)\left(\frac{14+6\sqrt{5}}{4}\right) \\ ={\left(\frac{28}{4}\right)}^2-2\left(\frac{{\left(14\right)}^2-{\left(6\sqrt{5}\right)}^2}{16}\right) \\ ={\left(7\right)}^2-2\left(\frac{196-180}{16}\right) \\ =49-2\left(\frac{16}{16}\right) \\ =49-2 \\ =47 \end{gathered}$$

Hence, the value of p² + q² is 47.

Question 22:

Rationalise the denominator of each of the following.
(i) $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$                 (ii) $\frac{3}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{5}}{\displaystyle -}{\displaystyle \sqrt{2}}}$             (iii) $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

Answer 22:

$$\begin{gathered} \left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}} \\ =\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}\times \frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}} \\ =\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^2-{\left(\sqrt{13}\right)}^2} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^2+{\left(\sqrt{6}\right)}^2+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^2} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\times \frac{\sqrt{42}}{\sqrt{42}} \\ =\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84} \\ =\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84} \end{gathered}$$
Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$.
$$\begin{gathered} \left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}} \\ =\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}\times \frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}} \\ =\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^2} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}} \\ =\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12} \\ =\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4} \end{gathered}$$
Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$.
$$\begin{gathered} \left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}} \\ =\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}\times \frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^2-{\left(\sqrt{7}\right)}^2} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^2+{\left(\sqrt{3}\right)}^2+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^2} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}} \\ =\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}} \\ =\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{2\sqrt{3}+3-\sqrt{21}}{3} \end{gathered}$$
Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.

Question 23:

Given, $\sqrt{2}=1.414 \mathrm{and} \sqrt{6}=2.449$, find the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal.

Answer 23:

$$\begin{gathered} \frac{1}{\sqrt{3}-\sqrt{2}-1} \\ =\frac{1}{\sqrt{3}-\left(\sqrt{2}+1\right)}\times \frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{\sqrt{3}+\left(\sqrt{2}+1\right)} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}+1\right)}^2} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2-2\left(\sqrt{2}\right)\left(1\right)-{\left(1\right)}^2} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{3-2-2\sqrt{2}-1} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{-2\sqrt{2}} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{-2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\sqrt{6}+2+\sqrt{2}}{-4} \\ =\frac{2.449+2+1.414}{-4} \left(\because \sqrt{2}=1.414 \mathrm{and} \sqrt{6}=2.449\right) \\ =\frac{5.863}{-4} \\ =-1.465 \end{gathered}$$

Hence, the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal is −1.465.

Question 24:

If $x=\frac{1}{2-\sqrt{3}}$, find the value of x³ – 2x² – 7x + 5.

Answer 24:

$$\begin{gathered} x=\frac{1}{2-\sqrt{3}} \\ \Rightarrow x=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\ \Rightarrow x=\frac{2+\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow x=\frac{2+\sqrt{3}}{4-3} \\ \Rightarrow x=2+\sqrt{3} ...\left(1\right) \\ \mathrm{Now}, \\ {x}^2={\left(2+\sqrt{3}\right)}^2 \\ \Rightarrow x^2={\left(2\right)}^2+{\left(\sqrt{3}\right)}^2+2\left(2\right)\left(\sqrt{3}\right) \\ \Rightarrow x^2=4+3+4\sqrt{3} \\ \Rightarrow x^2=7+4\sqrt{3} ...\left(2\right) \\ \mathrm{Also}, \\ {x}^3=x^2.x \\ \Rightarrow x^3=\left(7+4\sqrt{3}\right)\left(2+\sqrt{3}\right) \\ \Rightarrow x^3=14+7\sqrt{3}+8\sqrt{3}+12 \\ \Rightarrow x^3=26+15\sqrt{3} ...\left(3\right) \end{gathered}$$

Now,
$$\begin{gathered} x^3-2x^2-7x+5 \\ =\left(26+15\sqrt{3}\right)-2\left(7+4\sqrt{3}\right)-7\left(2+\sqrt{3}\right)+5 (\mathrm{using} \left(1\right), \left(2\right) \mathrm{and} \left(3\right)) \\ =26+15\sqrt{3}-14-8\sqrt{3}-14-7\sqrt{3}+5 \\ =31-28+15\sqrt{3}-15\sqrt{3} \\ =3 \end{gathered}$$

Hence, the value of x³ – 2x² – 7x + 5 is 3.

Question 25:

Evaluate $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$, it being given that $\sqrt{5}=2.236 \mathrm{and} \sqrt{10}=3.162$.
Hint
$\begin{array}{rcl}\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}& =& \frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}\\ & =& \frac{15}{3\sqrt{10}-3\sqrt{5}}=\frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\end{array}$

Answer 25:

$\begin{array}{rcl}\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}& =& \frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}\\ & =& \frac{15}{3\sqrt{10}-3\sqrt{5}}\\ & =& \frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\\ & =& \frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{{\left(\sqrt{10}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{5}\right)}^2}}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{10-5}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{5}\\ & =& \sqrt{10}+\sqrt{5}\\ & =& 3.162+2.236 \left(\mathrm{given}\right)\\ & =& 5.398\end{array}$

Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ = 5.398 .