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

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 for3+\sqrt2   is3-\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 for5\sqrt3-3\sqrt5   is 5\sqrt3+3\sqrt5   . We will multiply numerator and denominator of the given expression\frac{30}{5\sqrt3-3\sqrt5}   by5\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 to5\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 for2\sqrt2-\sqrt3is2\sqrt2+\sqrt3   . We will multiply numerator and denominator of the given expression\frac{\sqrt3+1}{2\sqrt2-\sqrt3}   by2\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 for6+4\sqrt2is6-4\sqrt2   . We will multiply numerator and denominator of the given expression\frac{6-4\sqrt2}{6+4\sqrt2}   by6-4\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 to17-12\sqrt2

(viii) We know that rationalisation factor for2\sqrt5-3   is2\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 to5-2\sqrt6   .

(ii) We know that rationalisation factor for7+4\sqrt3   is7-4\sqrt3   . We will multiply numerator and denominator of the given expression\frac{5+2\sqrt3}{7+4\sqrt3}   by7-4\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 to11-6\sqrt3   .

(iii) We know that rationalisation factor for3-2\sqrt2   is3+2\sqrt2   . We will multiply numerator and denominator of the given expression\frac{1+\sqrt2}{3-2\sqrt2}   by3+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 to7+5\sqrt2   .

(iv) We know that rationalisation factor for3\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}   by3\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 for2\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 for3\sqrt2+2\sqrt3   and\sqrt3-\sqrt2   are3\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 for3+\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 for2+\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.