Question

hello ,can u plz solve this for me​

Answer 1

Agar aapko lagta hai ki mera answer is worthy of being selected as the Brainliest then please GO aHEAD and

@BRAINLIEST

Answer 2

$since \: {(a + b)}^{2} = {a}^{2} + {b }^{2} + 2ab \\ = > {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab$

Therefore,

${x}^{2} + \frac{1}{ {x}^{2} } = ( \frac{ \sqrt{5} + 3 }{2} + \frac{2}{ \sqrt{5} + 3} ) ^{2} - 2 \times \frac{ \sqrt{5} + 3 }{2} \times \frac{2}{ \sqrt{5} + 3 } \\ = \frac{ {( \sqrt{5 } + 3)}^{2} + 4 }{2( \sqrt{5} + 3) } - 2$

$= \frac{5 + 9 + 6 \sqrt{5} + 4 }{2 \sqrt{5} + 6 } - 2 \\ \frac{18 + 6\sqrt{5} - 4 \sqrt{5} - 12}{2 \sqrt{5} +6} \\ = \frac{6 + 2 \sqrt{5} }{6 + 2 \sqrt{5} } \\ {x}^{2} + \frac{1}{ {x}^{2} } = 1$

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Thanks.