Question

please solve and give me answer as soon as possible

Answer 1

$\bold{\huge{\underline{\underline{Solution\: \colon}}}}$

$1) \:\dfrac{\sqrt{a+x} +\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}\\\\\\ Rationalising \: the\: denominator \: and \: numerator \: by \colon\\\\\\ multiply \: it \: by\\\\\\\sqrt{a+x} +\sqrt{a-x}\\\\\\ = \dfrac{\sqrt{a+x} +\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}} \times \dfrac{\sqrt{a+x}+ \sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\\\\\\= \dfrac{(\sqrt{a+x} +\sqrt{a-x})^{2}}{(\sqrt{a+x})^{2}-(\sqrt{a-x})^{2}}\\\\\\ = \dfrac{a+x+a-x+2(\sqrt{a+x}\times\sqrt{a-x})}{a+x-a+x}\\\\\\ = \dfrac{2a+2\sqrt{a^{2}-x^{2}}}{2x}\\\\\\ = \dfrac{2(a+\sqrt{a-x})}{2x}\\\\\\ = \dfrac{a+\sqrt{a^{2}-x^{2}}}{x}$

Now we have a value of x :

$x= \dfrac{2ab}{b^{2}+1}$

$= \dfrac{a+\sqrt{a^{2}-\Bigg(\dfrac{4a^{2}.b^{2}}{(b^{2}+1)^{2}}\Bigg)}}{\dfrac{2ab}{b^{2}+1}}\\\\\\= \dfrac{a+\sqrt{a^{2}\Bigg(1-\dfrac{4b^{2}}{(b^{2}+1)^{2}}\Bigg)}}{\dfrac{2ab}{b^{2}+1}}\\\\\\ = \dfrac{a+a\sqrt{\Bigg(1-\dfrac{4b^{2}}{(b^{2}+1)^{2}}\Bigg)}}{\dfrac{2ab}{b^{2}+1}}$

$= \dfrac{a\Bigg(1 + \sqrt{ \dfrac{ {( {b}^{2} + 1)}^{2} - 4 {b}^{2} }{ {( {b}^{2} + 1) }^{2} } }\Bigg)}{ \dfrac{2ab}{ {b}^{2} + 1} }$

$= \dfrac{a\Bigg(1 + \sqrt{ \dfrac{ ({b}^{4} + 2 {b}^{2} + 1 - 4 {b}^{2}) }{ {( {b}^{2} + 1) }^{2} } } \Bigg)}{ \dfrac{2ab}{ {b}^{2} + 1} }$

$= \dfrac{a\Bigg( \dfrac{ {b}^{2} + 1 + \sqrt{ {( {b}^{2} - 1) }^{2} } }{ {b}^{2} + 1}\Bigg)}{ \dfrac{2ab}{ {b}^{2} + 1} }$

$= \dfrac{a\Bigg( \dfrac{ {b}^{2} + \cancel{1} + {b}^{2} - \cancel{1} }{ \cancel{ {b}^{2} + 1}}\Bigg)}{ \dfrac{2ab}{ \cancel{{b}^{2} + 1} }}$

$= \dfrac{ \cancel{(2a b)}b}{ \cancel{(2ab)}} \\\\\\\\ = \bold{b}$

$\bold{\large{\boxed{Answer\: is \:(b)}}}$

$\bold{\large{Thanks}}$

Answer 2

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