Question

Please let me know this solution ...Thank You! ​

Answer 1

Step-by-step explanation:

$a = \dfrac{ \sqrt{5} + 1}{ \sqrt{5} - 1 } \\ = \dfrac{( \sqrt{5} + 1)( \sqrt{5} + 1) }{( \sqrt{5} - 1)( \sqrt{5} + 1) } \\ = \dfrac{ {( \sqrt{5} + 1) }^{2} }{ { (\sqrt{5}) }^{2} - {(1)}^{2} } \\ = \dfrac{ { (\sqrt{5}) }^{2} + {(1)}^{2} + 2( \sqrt{5} )(1) }{5 - 1} \\ = \dfrac{5 + 1 + 2 \sqrt{5} }{4} \\ = \dfrac{6 + 2 \sqrt{5} }{4} \\ = \dfrac{2(3 + \sqrt{5}) }{4} \\ a = \dfrac{3 + \sqrt{5} }{2} \\ similarly \\ b = \dfrac{ \sqrt{5} - 1}{ \sqrt{5} + 1 } \\ b = \dfrac{3 - \sqrt{5} }{2} \\ {a}^{2} = {( \frac{3 + \sqrt{5} }{2}) }^{2} \\ = \dfrac{9 + 5 + 2 \sqrt{5} }{4} \\ = \dfrac{14 +2 \sqrt{5} }{4} \\ similarly \\ {b}^{2} = \dfrac{14 - 2 \sqrt{5} }{4} \\ ab = ( \dfrac{3 + \sqrt{5} }{2} ) \: ( \dfrac{3 - \sqrt{5} }{2} ) \\ \\ = \dfrac{ {(3)}^{2} - { (\sqrt{5} )}^{2} }{4} \\ \\ = \dfrac{9 - 5}{4} \\ ab = \dfrac{4}{4} = 1$

${a}^{2} + {b}^{2} = {a}^{2} + {b}^{2} \\ \\ = \dfrac{14 + 2 \sqrt{5} }{4} + \dfrac{14 - 2 \sqrt{5} }{4} \\ \\ = \dfrac{14 + 2 \sqrt{5} + 14 - 2 \sqrt{5} }{4} \\ \\ = \dfrac{28}{4} \\ \\ = 7 \\ now \\ \\ \dfrac{ {a}^{2} + ab + {b}^{2} }{ {a}^{2} - ab + {b}^{2} } \\ \\ = \dfrac{ ({a}^{2} + {b}^{2} )+ ab } {( {a}^{2} + {b}^{2} )- ab } \\ \\ = \dfrac{7 + 1}{7 - 1} \\ \\ = \dfrac{8}{6} \\ \\ = \dfrac{4}{3}$