Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 16 May, 2021

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 on 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.

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :