Question

Simplify -
$\bf \: \dfrac{3}{ \sqrt{5} \: + \sqrt{2}} \: - \: \dfrac{2}{ \sqrt{7} \: + \sqrt{5}} \: + \: \dfrac{5}{ \sqrt{7} \: - \: \sqrt{2} }$
​

Answer 1

Answer:

$\bf \: \dfrac{3}{ \sqrt{5} \: + \sqrt{2}} \: - \: \dfrac{2}{ \sqrt{7} \: + \sqrt{5}} \: + \: \dfrac{5}{ \sqrt{7} \: - \: \sqrt{2} }$

First of all we will rationalise each fractions separately.

$\frac{3}{ \sqrt{5} + \sqrt{2} } \\ \\ multiplying \: the \: numerator \: and \: denominator \: by \: the \: conjugate \: of \: denominator$

$= \frac{3}{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \\ \\ = \frac{3( \sqrt{5} - \sqrt{2}) }{ {( \sqrt{5} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 ( \sqrt{5} - \sqrt{2} )}{5 - 2} \\ \\ = \frac{3( \sqrt{5} - \sqrt{2} )}{3} \\ \\ \large{= \sqrt{5} - \sqrt{2} }$

$\frac{2}{ \sqrt{7} + \sqrt{5} } \\ \\ rationalising \: the \: denominator$

$= \frac{2}{ \sqrt{7} + \sqrt{5} } \times \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \\ \\ = \frac{2( \sqrt{7} - \sqrt{5} )}{ {( \sqrt{7}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ = \frac{2( \sqrt{7} - \sqrt{5} )}{7 - 5} \\ \\ \frac{2( \sqrt{7} - \sqrt{5}) }{2} \\ \\ \large{= \sqrt{7} - \sqrt{5} }$

$\frac{5}{ \sqrt{7} - \sqrt{2} } \\ \\ rationalizing \: the \: denominator$

$= \frac{5}{ \sqrt{7} - \sqrt{2} } \times \frac{ \sqrt{7} + \sqrt{2} }{ \sqrt{7} + \sqrt{2} } \\ \\ = \frac{5( \sqrt{7} + \sqrt{2} )}{ {( \sqrt{7} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{5( \sqrt{7} + \sqrt{2} )}{7 - 2} \\ \\ = \frac{5( \sqrt{7} + \sqrt{2} )}{5} \\ \\ = \sqrt{7} + \sqrt{2}$

Now putting the rationalised fractions in

$\bf \: \dfrac{3}{ \sqrt{5} \: + \sqrt{2}} \: - \: \dfrac{2}{ \sqrt{7} \: + \sqrt{5}} \: + \: \dfrac{5}{ \sqrt{7} \: - \: \sqrt{2} }$

We get;

$= ( \sqrt{5} - \sqrt{2} ) - ( \sqrt{7} - \sqrt{5} ) + ( \sqrt{7} + \sqrt{2} ) \\ \\ = \sqrt{5} \: \: { \cancel{ - \sqrt{2} }} \: \: \cancel{ - \sqrt{7} }+ \sqrt{5} + \: \: \cancel{ \sqrt{7} }+ \: \: \cancel{\sqrt{2} } \\ \\ \\ = \sqrt{5} + \sqrt{5} \\ \\ \\ { \large{ \blue{ = 2 \sqrt{5} }}}$

$\huge{thus}$

${ \huge{ \boxed{\bf \: \dfrac{3}{ \sqrt{5} \: + \sqrt{2}} \: - \: \dfrac{2}{ \sqrt{7} \: + \sqrt{5}} \: + \: \dfrac{5}{ \sqrt{7} \: - \: \sqrt{2} } = 2 \sqrt{5} }}}$

Answer 2

Question:-

To simplify:-

• $\sf{\dfrac{3}{\sqrt{5} + \sqrt{2}} - \dfrac{2}{\sqrt{7} + \sqrt{5}} + \dfrac{5}{\sqrt{7} - \sqrt{2}}}$

Solution:-

$\sf{\dfrac{3}{\sqrt{5} + \sqrt{2}} - \dfrac{2}{\sqrt{7} + \sqrt{5}} + \dfrac{5}{\sqrt{7} - \sqrt{2}}\longrightarrow{(i)}}$

Let us rationalize all the terms separately,

$\sf{For,\: \dfrac{3}{\sqrt{5} + \sqrt{2}}}$

By rationalizing the denominator,

$=\sf{\dfrac{3}{\sqrt{5} + \sqrt{2}} \times \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}}$

$=\sf{\dfrac{3(\sqrt{5} - \sqrt{2})}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})}}$

Applying the identity in the denominator:-

$\boxed{\underline{\red{\rm{(a + b)(a - b) = a^2 - b^2}}}}$

$=\sf{\dfrac{3(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2}}$

$=\sf{\dfrac{3(\sqrt{5} - \sqrt{2})}{5 - 2}}$

$=\sf{\dfrac{3(\sqrt{5} - \sqrt{2})}{3}}$

$=\sf{\dfrac{\not{3}(\sqrt{5} - \sqrt{2})}{\not{3}}}$

$\blue{\boxed{\underline{\pink{\rm{\therefore\:\dfrac{3}{\sqrt{5} + \sqrt{2}} = \sqrt{5} - \sqrt{2}}}}}}$

$\sf{For,\:\dfrac{2}{\sqrt{7} + \sqrt{5}}}$

By rationalizing the denominator,

$=\sf{\dfrac{2}{\sqrt{7} + \sqrt{5}}\times \dfrac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}}}$

$=\sf{\dfrac{2(\sqrt{7} - \sqrt{5})}{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})}}$

$=\sf{\dfrac{2(\sqrt{7} - \sqrt{5})}{(\sqrt{7})^2 - (\sqrt{5})^2}}$

$=\sf{\dfrac{2(\sqrt{7} - \sqrt{5})}{7 - 5}}$

$=\sf{\dfrac{2(\sqrt{7} - \sqrt{5})}{2}}$

$=\sf{\dfrac{\not{2}(\sqrt{7} - \sqrt{5})}{\not{2}}}$

$\red{\boxed{\underline{\orange{\rm{\therefore\:\dfrac{2}{\sqrt{7} + \sqrt{5}} = \sqrt{7} - \sqrt{5}}}}}}$

$\sf{For,\:\dfrac{5}{\sqrt{7} - \sqrt{2}}}$

By rationalizing the denominator,

$=\sf{\dfrac{5}{\sqrt{7} - \sqrt{2}}\times \dfrac{\sqrt{7} + \sqrt{2}}{\sqrt{7} + \sqrt{7}}}$

$=\sf{\dfrac{5(\sqrt{7} + \sqrt{2})}{(\sqrt{7} - \sqrt{2})(\sqrt{7} + \sqrt{2})}}$

$=\sf{\dfrac{5(\sqrt{7} + \sqrt{2})}{(\sqrt{7})^2 - (\sqrt{2})^2}}$

$=\sf{\dfrac{5(\sqrt{7} + \sqrt{2})}{7 - 2}}$

$=\sf{\dfrac{5(\sqrt{7} + \sqrt{2})}{5}}$

$=\sf{\dfrac{\not{5}(\sqrt{7} + \sqrt{2})}{\not{5}}}$

$\purple{\boxed{\underline{\green{\rm{\therefore\:\dfrac{5}{\sqrt{7} - \sqrt{2}} = \sqrt{7} + \sqrt{2}}}}}}$

Putting all the values in (i)

= (√5 - √2) - (√7 - √5) + (√7 + √2)

= √5 - √2 - √7 + √5 + √7 + √2

= √5 + √5 - √2 + √2 - √7 + √7

= 2√5

$\boxed{\underline{\rm{\red{\therefore\:\dfrac{3}{\sqrt{5} + \sqrt{2}} - \dfrac{2}{\sqrt{7} + \sqrt{5}} + \dfrac{5}{\sqrt{7} - \sqrt{2}} = 2\sqrt{5}}}}}$

______________________________________