Class 9 RD Sharma Solutions Chapter 3 Rationalisation Exercise 3.2 | Set 1 (original) (raw)

Last Updated : 6 Feb, 2023

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.