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Answer»

\small{\underline{\sf{Question:-}}}

\small{ \dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}}

\large{\underline{\sf{Solution:-}}}

Consider,

  • The required answer be x

= \dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}} = x

Now,

= {\dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}}^{2} = x²

Using identity,

  • (a + b)² = a² + 2ab + b²

= \small{\dfrac{\sqrt{5} - 2 + 2 \sqrt{(\sqrt{5} - 2) × (\sqrt{5} + 2)} + \sqrt{5} + 2}{\sqrt{5} + 1}} = \sf{ {x}^{2}}

= {\dfrac{\sqrt{5} \cancel{- 2} + 2 \sqrt{(\sqrt{5} - 2) × (\sqrt{5} + 2)} + \sqrt{5} \cancel{+ 2} } {\sqrt{5} + 1}} = \sf{ {x}^{2}}

= \dfrac{\sqrt{5} + 2 \sqrt{ 5 - 4} + \sqrt{5}}{\sqrt{5} + 1} = x²

= \dfrac{2 \sqrt{5} + 2 \sqrt{ 1 }} {\sqrt{5} + 1} = x²

Now,

  • Take 2 as common and factorize

= \dfrac{2( \sqrt{5} + 1 ) } {\sqrt{5} + 1} = x²

= \dfrac{2 \cancel{( \sqrt{5} + 1 ) }} {\cancel{\sqrt{5} + 1}} = x²

= 2 = x²

= √2 = x

Thus, the required answer is √2 respectively!!

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