Class 9 RD Sharma Solutions – Chapter 3 Rationalisation- Exercise 3.2 | Set 1

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Question 1. Rationalise the denominator of each of the following (i-vii):

(i) $\frac{3}{\sqrt5}$

(ii) $\frac{3}{2\sqrt5}$

(iii) $\frac{1}{\sqrt{12}}$

(iv) $\frac{\sqrt3}{\sqrt5}$

(v) $\frac{\sqrt3+1}{\sqrt2}$

(vi) $\frac{\sqrt2+\sqrt5}{3}$

(vii) $\frac{3\sqrt{2}}{\sqrt5}$

Solution:

(i) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{3}{\sqrt5}$ by $\sqrt5$ . to

get $\frac{3}{\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{3\sqrt5}{\sqrt5\times\sqrt5}\\ \frac{3\sqrt5}{\sqrt5}$

Hence, the given expression is simplified to $\frac{3\sqrt5}{5}$ .

(ii) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{3}{2\sqrt5}$ by $\sqrt5$ . to

get $\frac{3}{2\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{3\sqrt5}{2\sqrt5+\sqrt5}\\ =\frac{3\sqrt5}{2\times5}\\ =\frac{3\sqrt5}{10}$

Hence, the given expression is simplified to $\frac{3\sqrt5}{10}$

(iii) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{1}{\sqrt{12}}$ by $\sqrt{12}$ .

to get $\frac{1}{\sqrt{12}}\times\frac{\sqrt{12}}{\sqrt{12}}=\frac{\sqrt{12}}{\sqrt{12}+\sqrt{12}}\\ =\frac{\sqrt{12}}{12}\\ =\frac{\sqrt{14}\times\sqrt3}{12}\\ =\frac{2\times\sqrt3}{12}\\ =\frac{\sqrt3}{6}$

Hence the given expression is simplified to $\frac{\sqrt3}{6}$

(iv) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt2}{\sqrt5}$ is $\sqrt5$ .

to get $\frac{\sqrt2}{\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{\sqrt2\times\sqrt5}{\sqrt5\times\sqrt5}\\ \frac{\sqrt{10}}{5}$

Hence, the given expression is simplified to $\frac{\sqrt{10}}{5}$

(v) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt3+1}{\sqrt2}$ by $\sqrt2$ to get $\frac{\sqrt3+1}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2\times\sqrt3+\sqrt2}{\sqrt2\times\sqrt2}\\ =\frac{\sqrt6+\sqrt2}{2}$

Hence, the given expression is simplified to $\frac{\sqrt6+\sqrt2}{2}$

(vi) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt2+\sqrt5}{\sqrt3}$ by $\sqrt3$ to get $\frac{\sqrt2+\sqrt5}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{\sqrt2\times\sqrt3+\sqrt5\times\sqrt3}{\sqrt3\times\sqrt3}\\ \frac{\sqrt6+\sqrt{15}}{3}$

Hence, the given expression is simplified to $\frac{\sqrt6+\sqrt{15}}{3}$ .

(vii) We know that rationalisation factor for $\frac{1}{\sqrt{a}}$ is $\sqrt{a}$ . We will multiply numerator and denominator of the given expression $\frac{3\sqrt2}{\sqrt5}$ by $\sqrt5$ to get $\frac{3\sqrt2}{\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{3\sqrt2\times\sqrt5}{\sqrt5\times\sqrt5}\\ =\frac{3\sqrt{10}}{5}$

Hence, the given expression is simplified to $\frac{3\sqrt{10}}{5}$

Question 2. Find the value to three places of decimals of each of the following. It is given that $\sqrt2-1.414,\ \ \sqrt3-1.732,\ \ \sqrt5-2.236\ and\ \sqrt{10}-3.162$

(i) $\frac{2}{\sqrt3}$

(ii) $\frac{3}{\sqrt{10}}$

(iii) $\frac{\sqrt5+1}{\sqrt2}$

(iv) $\frac{\sqrt{10}+\sqrt{15}}{\sqrt2}$

(v) $\frac{2+\sqrt3}{3}$

(vi) $\frac{\sqrt2-1}{\sqrt5}$

Solution:

(i) We know that rationalisation factor of the denominator is $\sqrt3$ . We will multiply numerator and denominator of the given expression $\frac{2}{\sqrt3}$ by $\sqrt3$ to get

$\frac{2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{2\times\sqrt3}{\sqrt3+\sqrt3}\\ =\frac{2\sqrt3}{3}\\ =\frac{2\times1.732}{3}\\ =\frac{3.4641}{3}\\ =1.1547$

The value of expression 1.1547 can be round off to decimal places as 1.155.

Hence, the given expression is simplified to 1.155.

(ii) We know that rationalisation factor of the denominator is $\sqrt{10}$ . We will multiply numerator and denominator of the given expression $\frac{3}{\sqrt{10}}by\sqrt{10}$

to get

$\frac{3}{\sqrt{10}}\times\frac{\sqrt{10}}{\sqrt{10}}=\frac{3\times\sqrt{10}}{\sqrt{10}\times\sqrt{10}}\\ =\frac{3\sqrt{10}}{10}\\ =\frac{3\times3.162}{10}\\ \frac{9.486}{10}\\ =0.9486$

The value of expression 0.9486 can be round off to decimal places as 0.949.

Hence, the given expression is simplified to 0.949.

(iii) We know that rationalisation factor of the denominator is $\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt5+1}{\sqrt2}$ by $\sqrt2$

to get

$\frac{\sqrt5+1}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{10}+\sqrt2}{\sqrt2\times\sqrt2}\\ =\frac{\sqrt{10}+\sqrt2}{2}\\ =\frac{3.162+1.414}{2}\\ =\frac{4.576}{2}\\ =2.288$

The value of expression 2.288 can be round off to decimal places as 2.288.

Hence, the given expression is simplified to 2.288.

(iv) We know that rationalisation factor of the denominator is $\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt{10}+\sqrt{15}}{\sqrt2}$

by $\sqrt2$ to get

$\frac{\sqrt{10}+\sqrt{15}}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{10}\times\sqrt2+\sqrt{15}\times\sqrt2}{\sqrt2\times\sqrt2}\\ =\frac{\sqrt{10}\times\sqrt2+\sqrt5\times\sqrt3\times\sqrt2}{2}\\ =\frac{3.162\times1.414+2.236\times1.732\times1.414}{2}\\ =\frac{9.947}{2}\\ =4.9746$

The value of expression 4.9746 can be round off to decimal places as 4.975.

Hence, the given expression is simplified to 4.975.

(v) We know that rationalisation factor of the denominator is $\sqrt3$ . We will multiply numerator and denominator of the given expression $\frac{2+\sqrt3}{2}by\sqrt3$ to get

$\frac{2+\sqrt3}{2}=\frac{2+1.732}{2}\\ =\frac{3.732}{2}\\ =1.24401$

The value of expression 1.24401 can be round off to decimal places as 1.244.

Hence, the given expression is simplified to 1.244.

(vi) We know that rationalisation factor of the denominator is $\sqrt5$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt2-1}{\sqrt5}by \sqrt5$

to get

$\frac{\sqrt2-1}{\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{\sqrt2\times\sqrt5-\sqrt5}{\sqrt5\times\sqrt5}\\ =\frac{\sqrt{10}-\sqrt5}{5}$

Putting the value of $\sqrt{10}$ and $\sqrt5$ , we get

The value of expression 0.1852 can be round off to decimal places as 0.185.

Hence, the given expression is simplified to 0.185

Question 3. Express each one of the following with rational denominator:

(i) $\frac{1}{3+\sqrt2}$

(ii) $\frac{1}{\sqrt6-\sqrt5}$

(iii) $\frac{16}{\sqrt{41}-5}$

(iv) $\frac{30}{5\sqrt3-3\sqrt5}$

(v) $\frac{1}{2\sqrt5-\sqrt3}$

(vi) $\frac{\sqrt3+1}{2\sqrt2-\sqrt3}$

(vii) $\frac{6-4\sqrt2}{6+4\sqrt2}$

(viii) $\frac{3\sqrt2+1}{2\sqrt5-3}$

(ix) $\frac{b^2}{\sqrt{a^2+b^2}+a^2}$

Solution:

(i) We know that rationalisation factor for $3+\sqrt2$ is $3-\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{1}{3+\sqrt2}by3-\sqrt2$ to get

$\frac{1}{3+\sqrt2}\times\frac{3-\sqrt2}{3-\sqrt2}=\frac{3-\sqrt2}{3^2-(\sqrt2)^2}\\ =\frac{3-\sqrt2}{9-2}\\ =\frac{3-\sqrt2}{7}$

Hence, the given expression is simplified with rational denominator to $\frac{3-\sqrt2}{7}$

(ii) We know that rationalisation factor for $\sqrt6-\sqrt5$ is $\sqrt6+\sqrt5$ . We will multiply numerator and denominator of the given expression $\frac{1}{\sqrt6-\sqrt5}$ by $\sqrt6+\sqrt5$

to get

$\frac{1}{\sqrt6-\sqrt5}\times\frac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}=\frac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}\\ =\frac{\sqrt6+\sqrt5}{6-5}\\ =\frac{\sqrt6+\sqrt5}{1}\\ =\sqrt6+\sqrt5$

Hence, the given expression is simplified with rational denominator to $\sqrt6+\sqrt5$

(iii) We know that rationalisation factor for $\sqrt{41}-5$ is $\sqrt{41}+5$ . We will multiply numerator and denominator of the given expression $\frac{16}{\sqrt{41}-5}$ by $\sqrt{41}+5$

to get

$\frac{16}{\sqrt{41}-5}\times\frac{\sqrt{41}+5}{\sqrt{41}+5}=\frac{16(\sqrt{41}+5)}{(\sqrt{41})^2-(\sqrt{5})^2}\\ =\frac{16(\sqrt{41}+5)}{41-25}\\ =\frac{16(\sqrt{41}+5)}{16}\\ =\sqrt{41}+5$

Hence, the given expression is simplified with rational denominator to $\sqrt{41}+5$

(iv) We know that rationalisation factor for $5\sqrt3-3\sqrt5$ is $5\sqrt3+3\sqrt5$ . We will multiply numerator and denominator of the given expression $\frac{30}{5\sqrt3-3\sqrt5}$ by $5\sqrt3+3\sqrt5$

to get

$\frac{30}{5\sqrt3-3\sqrt5}\times\frac{5\sqrt3+3\sqrt5}{5\sqrt3+3\sqrt5}=\frac{30\times5\times\sqrt3+30\times3\times\sqrt5}{25\times3-9\times5}\\ =\frac{30\times5\times\sqrt3+30\times3\times\sqrt5}{25\times3-9\times5}\\ =\frac{30\times5\times\sqrt3+30\times3\times\sqrt5}{75-45}\\ =\frac{30\times5\times\sqrt3+30\times3\times\sqrt5}{30}\\ =5\sqrt3+3\sqrt5$

Hence, the given expression is simplified with rational denominator to $5\sqrt3+3\sqrt5$

(v) We know that rationalisation factor for is . We will multiply numerator and denominator of the given expression by to get

$\frac{1}{2\sqrt5-\sqrt3}\times\frac{2\sqrt5+\sqrt3}{2\sqrt5+\sqrt3}=\frac{2\sqrt5+\sqrt3}{(2\sqrt5)^2-(\sqrt3)^2}\\ =\frac{2\sqrt5+\sqrt3}{4\times5-3}\\ =\frac{2\sqrt5+\sqrt3}{20-3}\\ =\frac{2\sqrt5+\sqrt3}{17}\\ =5\sqrt3+3\sqrt5$

Hence, the given expression is simplified with rational denominator to $\frac{2\sqrt5+\sqrt3}{17}$

(vi) We know that rationalisation factor for $2\sqrt2-\sqrt3is2\sqrt2+\sqrt3$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt3+1}{2\sqrt2-\sqrt3}$ by $2\sqrt2+\sqrt3$ to get

$\frac{\sqrt3+1}{2\sqrt2-\sqrt3}\times\frac{2\sqrt2+3}{2\sqrt2+\sqrt3}=\frac{2\times\sqrt3\times\sqrt2+\sqrt3\times\sqrt3+2\sqrt2+\sqrt3}{(2\sqrt2)^2-(\sqrt3)^2}\\ =\frac{2\sqrt{3\times2}+3+2\sqrt2+\sqrt3}{4\times2-3}\\ =\frac{2\sqrt6+3+2\sqrt2+\sqrt3}{8-3}\\ =\frac{2\sqrt6+3+2\sqrt2+\sqrt3}{5}\\ =\frac{2\sqrt6+3+2\sqrt2+\sqrt3}{5}$

Hence, the given expression is simplified with rational denominator to= $\frac{2\sqrt6+3+2\sqrt2+\sqrt3}{5}$

(vii) We know that rationalisation factor for $6+4\sqrt2is6-4\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{6-4\sqrt2}{6+4\sqrt2}$ by $6-4\sqrt2$ to get

$\frac{6-4\sqrt2}{6+4\sqrt2}\times\frac{6-4\sqrt2}{6-4\sqrt2}=\frac{6^2+(4\sqrt2)^2-2\times6\times4\sqrt2}{(6)^2-(4\sqrt2)^2}\\ =\frac{36+16\times2-48\sqrt2}{36-16\times2}\\ =\frac{68-48\sqrt2}{4}\\ =17-12\sqrt2$

Hence, the given expression is simplified with rational denominator to $17-12\sqrt2$

(viii) We know that rationalisation factor for $2\sqrt5-3$ is $2\sqrt5+3$ . We will multiply numerator and denominator of the given expression $\frac{3\sqrt2+1}{2\sqrt5-3}by2\sqrt5+3$ to get

$\frac{3\sqrt2+1}{2\sqrt5-3}\times\frac{2\sqrt5+3}{2\sqrt5+3}=\frac{3\sqrt2\times2\sqrt5+3\times3\sqrt2+2\sqrt5+3}{(2\sqrt5)^2-(3)^2}\\ =\frac{3\times2\times\sqrt2\times\sqrt5+\sqrt3\times3\sqrt2+2\sqrt5+3}{4\times5-9}\\ =\frac{6\sqrt{2\times5}+9\sqrt2+2\sqrt5+3}{4\times5-9}\\ =\frac{6\sqrt{10}+9\sqrt2+2\sqrt5+3}{11}$

Hence, the given expression is simplified with rational denominator to $\frac{6\sqrt{10}+9\sqrt2+2\sqrt5+3}{11}$

(ix) We know that rationalisation factor for $\sqrt{a^2+b^2}+a$ is $\sqrt{a^2+b^2}-a$ . We will multiply numerator and denominator of the given expression $\frac{b^2}{\sqrt{a^2+b^2}+a}$ by $\sqrt{a^2+b^2}-a$ to get

$\frac{b^2}{\sqrt{a^2+b^2}+a}\times\frac{\sqrt{a^2+b^2}-a}{\sqrt{a^2+b^2}-a}=\frac{b^2(\sqrt{a^2+b^2-a)}}{(\sqrt{a^2+b^2})^2-a^2}\\ =\frac{b^2(\sqrt{a^2+b^2}-a)}{a^2+b^2-a}\\ =\frac{b^2(\sqrt{a^2+b^2}-a)}{b^2}\\ =\sqrt{a^2+b^2}-a$

Hence, the given expression is simplified with rational denominator to $\sqrt{a^2+b^2}-a$

Question 4. Rationales the denominator and simplify:

(i) $\frac{3-\sqrt2}{3+\sqrt2}$

(ii) $\frac{5+2\sqrt3}{7+4\sqrt3}$

(iii) $\frac{1+\sqrt2}{3-2\sqrt2}$

(iv) $\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$

(v) $\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$

(vi) $\frac{2\sqrt3-\sqrt5}{2\sqrt2+3\sqrt3}$

Solution:

(i) We know that rationalisation factor for $\sqrt3+\sqrt2$ is $\sqrt3-\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$ by $\sqrt3-\sqrt2$ to get

$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}=\frac{(\sqrt3)^2+(2)^2-2\times\sqrt3\times\sqrt2}{(\sqrt3)^2-(\sqrt2)^2}\\ =\frac{3+2-2\sqrt6}{3-2}\\ =\frac{5-2\sqrt6}{1}\\ =5-2\sqrt6$

Hence, the given expression is simplified to $5-2\sqrt6$ .

(ii) We know that rationalisation factor for $7+4\sqrt3$ is $7-4\sqrt3$ . We will multiply numerator and denominator of the given expression $\frac{5+2\sqrt3}{7+4\sqrt3}$ by $7-4\sqrt3$ to get

$\frac{5+2\sqrt3}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}=\frac{5\times7-5\times4\sqrt3+2\times7\times\sqrt3-2\times4\times(\sqrt3)^2}{(7)^2-(4\sqrt3)^2}\\ =\frac{35-20\sqrt3+14\sqrt3-8\times3}{49-48}\\ =\frac{11-6\sqrt3}{1}\\ =11-6\sqrt3$

Hence, the given expression is simplified to $11-6\sqrt3$ .

(iii) We know that rationalisation factor for $3-2\sqrt2$ is $3+2\sqrt2$ . We will multiply numerator and denominator of the given expression $\frac{1+\sqrt2}{3-2\sqrt2}$ by $3+2\sqrt2$ to get

$\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=\frac{3+2\sqrt2+3\sqrt2\times(\sqrt2)^2}{(3)^2-(2\sqrt2)^2}\\ =\frac{3+5\sqrt2+4}{9-4\times2}\\ =\frac{7+5\sqrt2}{9-8}\\ =\frac{7+5\sqrt2}{1}\\ =7+5\sqrt2$

Hence, the given expression is simplified to $7+5\sqrt2$ .

(iv) We know that rationalisation factor for $3\sqrt5-2\sqrt6$ is $3\sqrt5+2\sqrt6$ . We will multiply numerator and denominator of the given expression $\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$ by $3\sqrt5+2\sqrt6$ to get

$\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}\times\frac{3\sqrt5+2\sqrt6}{3\sqrt5+2\sqrt6}=\frac{2\times3\times\sqrt6\times\sqrt5+(2\sqrt6)^2-3\times(\sqrt5)^2-2\times\sqrt5\times\sqrt6}{(3\sqrt5)^2-(2\sqrt6)^2}\\ =\frac{6\sqrt{6\times5}+4\times6-3\times5-2\times\sqrt{5\times6}}{9\times5-4\times6}\\ =\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}\\ =\frac{9+4\sqrt{30}}{21}$

Hence, the given expression is simplified to $\frac{9+4\sqrt{30}}{21}$ .

(v) We know that rationalisation factor for $\sqrt{48}+\sqrt{18}is\sqrt{48}-\sqrt{18}$ . We will multiply numerator and denominator of the given expression $\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$ by $\sqrt{48}-\sqrt{18}$ to get

$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}\times\frac{\sqrt{48}-\sqrt{18}}{\sqrt{48}-\sqrt{18}}=\frac{4\times\sqrt3\times\sqrt{48}-4\times\sqrt3\times\sqrt{18}+5\times\sqrt2\times\sqrt{48}-5\times\sqrt2\times\sqrt{18}}{(\sqrt{48})^2-(\sqrt{18})^2}\\ =\frac{4\sqrt{3\times48}-4\times\sqrt{3\times18}+5\times\sqrt{2\times48}-5\times\sqrt{2\times18}}{48-18}\\ =\frac{48-12\sqrt6+20\sqrt6-30}{30}\\ =\frac{18+8\sqrt6}{30}\\ =\frac{9+4\sqrt6}{15}$

Hence, the given expression is simplified to $\frac{9+4\sqrt6}{15}$ .

(vi) We know that rationalisation factor for $2\sqrt2+3\sqrt3$ is $2\sqrt2-3\sqrt3$ . We will multiply numerator and denominator of the given expression $\frac{2\sqrt3-\sqrt5}{2\sqrt2+3\sqrt3}by2\sqrt2-3\sqrt3$ to get

$\frac{2\sqrt3-\sqrt5}{2\sqrt2+3\sqrt3}\times\frac{2\sqrt2-3\sqrt3}{2\sqrt2-3\sqrt3}=\frac{2\times2\times\sqrt3\times\sqrt2-2\times3\times\sqrt3\times\sqrt3-2\times\sqrt5\times\sqrt2+3\times\sqrt5\times\sqrt3}{(2\sqrt2)^2-(3\sqrt3)^2}\\ =\frac{4\sqrt{3\times2}-6\times(\sqrt3)^2-2\times\sqrt{5\times2}+3\times\sqrt{5\times3}}{4\times2-9\times3}\\ =\frac{4\sqrt6-18-2\sqrt{10}+3\sqrt{15}}{-19}\\ =\frac{18+2\sqrt{10}-3\sqrt{15}-4\sqrt6}{19}$

Hence, the given expression is simplified to $\frac{18+2\sqrt{10}-3\sqrt{15}-4\sqrt6}{19}$ .

Question 5. Simplify:

(i) $\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}\\$

(ii) $\frac{5+\sqrt3}{5-\sqrt3}+\frac{5-\sqrt3}{5+\sqrt3}$

(iii) $\frac{7+3\sqrt5}{3+\sqrt5}+\frac{7-3\sqrt5}{3-\sqrt5}$

(iv) $\frac{1}{2+\sqrt3}+\frac{2}{\sqrt5-\sqrt3}+\frac{1}{2-\sqrt5}$

(v) $\frac{2}{\sqrt5+\sqrt3}+\frac{1}{\sqrt3+\sqrt2}+\frac{3}{\sqrt5+\sqrt2}$

Solution:

(i) We know that rationalisation factor for $3\sqrt2+2\sqrt3$ and $\sqrt3-\sqrt2$ are $3\sqrt2-2\sqrt3$ and $\sqrt3+\sqrt2$ respectively.

We will multiply numerator and denominator of the given expression $\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}\ and\ \frac{\sqrt{12}}{\sqrt3-\sqrt2}\ by\ 3\sqrt2-2\sqrt3\ and\ \sqrt3+\sqrt2$ respectively, to get

$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}\times\frac{3\sqrt2-2\sqrt3}{3\sqrt2-2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\frac{(3\sqrt2)^2+(2\sqrt3)^2-2\times3\sqrt2\times2\sqrt3}{(3\sqrt2)^2-(2\sqrt3)^2}+\frac{\sqrt{36}+\sqrt{24}}{(\sqrt3)^2-(\sqrt2)^2}\\ =\frac{18+12-12\sqrt6}{18-12}+\frac{6+\sqrt{24}}{3-2}\\ =\frac{30-12\sqrt6+36+12\sqrt6}{6}\\ =\frac{66}{6}\\ =11$

Hence, the given expression is simplified to 11.

(ii) We know that rationalisation factor for $\sqrt5-\sqrt3\ and\ \sqrt5+\sqrt3\ are\ \sqrt5+\sqrt3\ and\ \sqrt5-\sqrt3$ respectively.

We will multiply numerator and denominator of the given expression $\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\ and\ \frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\ by\ \sqrt5+\sqrt3\ and\ \sqrt5+\sqrt3$ respectively, to get

$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}=\frac{\sqrt5^2+\sqrt3^2+2\times\sqrt5\times\sqrt3}{\sqrt5^2-\sqrt3^2}+\frac{\sqrt5^2+\sqrt3^2-2\times\sqrt5\times\sqrt3}{\sqrt5^2-\sqrt3^2}\\ =\frac{5+3+2\sqrt{15}}{5-3}+\frac{5+3-2\sqrt{15}}{5-3}\\ =\frac{5+3+2\sqrt{15}+5+3-2\sqrt{15}}{2}\\ =\frac{16}{2}\\ =8$

Hence, the given expression is simplified to 8.

(iii) We know that rationalisation factor for $3+\sqrt5\ and\ 3-\sqrt5\ are\ 3-\sqrt5\ and\ 3+\sqrt5$ respectively.

We will multiply numerator and denominator of the given expression $\frac{7+3\sqrt5}{3+\sqrt5}\ and\ \frac{7-3\sqrt5}{3-\sqrt5}\ by\ 3-\sqrt5\ and\ 3+\sqrt5$ respectively, to get

$\frac{7+3\sqrt5}{3+\sqrt5}\times\frac{3-\sqrt5}{3-\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5}=\frac{7\times3-7\times\sqrt5+9\times\sqrt5-3\times\sqrt5^2}{3^2-\sqrt5^2}-\frac{7\times3+7\times\sqrt5-9\times\sqrt5-3\times\sqrt5^2}{3^2-\sqrt5^2}\\ =\frac{21-7\sqrt5+9\sqrt5-3\times5}{9-5}-\frac{21+7\sqrt5-9\sqrt5-3\times5}{9-5}\\ =\frac{6+2\sqrt5-6+2\sqrt5}{4}\\ =\frac{4\sqrt5}{4}\\ =\sqrt5$

Hence, the given expression is simplified to $\sqrt5$ .

(iv) We know that rationalisation factor for $2+\sqrt3,\ \sqrt5-\sqrt3\ and\ 2-\sqrt5\ are\ 2-\sqrt3,\ \sqrt5+\sqrt3\ and\ 2+\sqrt5$ respectively.

We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3},\ \frac{2}{\sqrt5-\sqrt3}\ and\ \frac{1}{2-\sqrt5}\ by\ 2-\sqrt3,\ \sqrt5+\sqrt3\ and\ 2+\sqrt5$ respectively, to get

$\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}+\frac{2}{\sqrt5+\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{1}{2-\sqrt5}\times\frac{2+\sqrt5}{2+\sqrt5}=\frac{2-\sqrt3}{2^2-\sqrt3^2}+\frac{2\sqrt5+2\sqrt3}{\sqrt5^2-\sqrt3^2}+\frac{2-\sqrt5}{2^2-\sqrt5^2}\\ =\frac{2-\sqrt3}{1}+\frac{2\sqrt5+2\sqrt3}{2}+\frac{2+\sqrt5}{-1}\\ =2-\sqrt3+\sqrt5+\sqrt3-\sqrt5-2\\ =0$

Hence, the given expression is simplified to 0 .

(v) We know that rationalisation factor for $\sqrt5+\sqrt3,\ \sqrt3+\sqrt2\ and\ \sqrt5+\sqrt2\ are\ \sqrt5-\sqrt3,\ \sqrt3-\sqrt2\ and\ \sqrt5-\sqrt2$ respectively.

We will multiply numerator and denominator of the given expression $\frac{2}{\sqrt5+\sqrt3},\ \frac{1}{\sqrt3+\sqrt2}\ and\ \frac{3}{\sqrt5+\sqrt2}\ by\ 2-\sqrt3,\ \sqrt5+\sqrt3\ and\ 2+\sqrt5$ respectively, to get

$\frac{2}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}+\frac{1}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}-\frac{3}{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}=\frac{2\sqrt5-2\sqrt3}{5-3}+\frac{\sqrt3-\sqrt2}{3-2}-\frac{3\sqrt5-3\sqrt2}{5-2}\\ =\frac{2\sqrt5-2\sqrt3}{2}+\frac{\sqrt3+\sqrt2}{1}-\frac{3\sqrt5-3\sqrt2}{3}\\ =\sqrt5-\sqrt3+\sqrt3-\sqrt2-\sqrt5+\sqrt2\\ =0$

Hence, the given expression is simplified to 0.