Question

$<b><big>Hello Friends$

$<b><big>Answer the above question$

$<b><big>No Spam$

Answer 1

$\dfrac{4}{2+\sqrt{3}-\sqrt{7}} = \sqrt{a} + \sqrt{b} +\sqrt{c}$  

By Rationalization: Multiply and divide by 2 + √3 - √7 on numerator and denominator both,  

$\dfrac{4 \times ( 2 + \sqrt{3} + \sqrt{7})}{( 2 +\sqrt{3} - \sqrt{3})(2 + \sqrt{3} + \sqrt{7} )}= \sqrt{a} + \sqrt{b} +\sqrt{c}$  

We know, ( a + b )( a - b ) = a^2 - b^2  

∴ $\dfrac{4( 2 + \sqrt{3} + \sqrt{7}}{ ( 2 + \sqrt{7} )^2 - ( \sqrt{7} )^2 }= \sqrt{a} + \sqrt{b} +\sqrt{c}$

We know, ( a + b )^2 = a^2 + b^2 + 2ab  

$\dfrac{4( 2 + \sqrt{3} +\sqrt{7}}{4 + 3 + 4\sqrt{7} - 7}=\sqrt{a} + \sqrt{b} +\sqrt{c}$  

$\dfrac{4(2+\sqrt{3}+\sqrt{7}}{4\sqrt{3}}=\sqrt{a} + \sqrt{b} +\sqrt{c}$  

$\dfrac{2+\sqrt{3}+\sqrt{7}}{\sqrt{3}}=\sqrt{a} + \sqrt{b} +\sqrt{c}$  

$\dfrac{2}{\sqrt{3}} +\dfrac{\sqrt{3}}{\sqrt{3}} +\dfrac{\sqrt{7}}{\sqrt{3}}=\sqrt{a} + \sqrt{b} +\sqrt{c}$

By Rationalization  

$\dfrac{2\sqrt{3}}{3} +1 + \dfrac{\sqrt{21}}{3}=\sqrt{a} + \sqrt{b} +\sqrt{c}$  

As all terms in left hand side and right hand side are positive, so sum of square terms on right hand side will be equal to the sum of square of terms in left hand side.

$\therefore \bigg( \dfrac{2\sqrt{3}}{3} \bigg)^2 + 1^2 + \bigg( \dfrac{\sqrt{21}}{3} \bigg)^2= (\sqrt{a})^2 + (\sqrt{b} )^2+(\sqrt{c})^2$  

$\dfrac{4}{3} + 1 + \dfrac{21}{9}$ = a + b + c

$\dfrac{12+9+21}{9}$ = a + b + c

$\dfrac{42}{9}$ = a + b + c

$\dfrac{14}{3}$ = a + b + c

Therefore, option D is correct.

Answer 2

Hello!

answer's in the attachment.
this is just a case where I've given explaination in detail.
you can always feel free to eliminate the extra calculation steps :)


Hope you get it :)