Question

answer this
it's too urgent
Please!!!

Answer 1

this is answer of this question

Answer 2

Hey friend,
Here is the answer you were looking for:
$a = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \\ \\ using \: the \: identities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{5} )}^{2} + {( \sqrt{3} )}^{2} - 2( \sqrt{5})( \sqrt{3} )}{ {( \sqrt{5}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{5 + 3 - 2 \sqrt{15} }{5 - 3} \\ \\ = \frac{8 - 2 \sqrt{15} }{2} \\ \\ = 4 - \sqrt{15} \\ \\ b = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{5} )}^{2} + {( \sqrt{3} )}^{2} + 2( \sqrt{5})( \sqrt{3} )}{ {( \sqrt{5}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{5 + 3 + 2 \sqrt{15} }{5 - 3} \\ \\ = \frac{8 + 2 \sqrt{15} }{2} \\ \\ = 4 + \sqrt{15} \\ \\a + b + ab \\ \\ = (4 - \sqrt{15} ) + (4 + \sqrt{15} ) + (4 - \sqrt{15} )(4 - \sqrt{15} ) \\ \\ = 4 - \sqrt{15} + 4 + \sqrt{15} + ( {(4)}^{2} - {( \sqrt{15}) }^{2} ) \\ \\ = 4 + 4 + (16 - 15) \\ \\ = 4 + 4 + 1 \\ \\ = 9$

Hope this helps!!!

@Mahak24

Thanks...
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