(((√((√5)+2)) + (√((√5)−2)))/(√((√5)+1))) ?


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Question Number 105018 by bemath last updated on 25/Jul/20

$\frac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}} ?$
Commented by bemath last updated on 25/Jul/20

$t h a n k y o u a l l$
	Answered by som(math1967) last updated on 25/Jul/20
	$let (((√((√5)+2))+(√((√5)−2)))/(√((√5)+1)))=p ⇒{(√((√5)+2))+(√((√5)−2))}^2 =p^2 ((√5)+1) (√5)+2+(√5)−2+2(√(5−2^2 ))=p^2 ((√5)+1{ 2(√5)+2=p^2 ((√5)+1) ∴p^2 =((2((√5)+1))/(((√5)+1)))=2 ∴p=(√2) ans$
	$let \frac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}} = p$ $\Rightarrow {\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}^{2} = p^{2} (\sqrt{5} + 1)$ $\sqrt{5} + 2 + \sqrt{5} - 2 + 2 \sqrt{5 - 2^{2}} = p^{2} (\sqrt{5} + 1 {$ $2 \sqrt{5} + 2 = p^{2} (\sqrt{5} + 1)$ $∴ p^{2} = \frac{2 (\sqrt{5} + 1)}{(\sqrt{5} + 1)} = 2$ $∴ p = \sqrt{2} ans$
	Answered by Dwaipayan Shikari last updated on 25/Jul/20
	$a^2 =(((√5)+2+(√5)−2+2(√(5−4)))/((√5)+1))=2(((√5)+1)/((√5)+1))=2 {take it as a} a=(√2)$
	$a^{2} = \frac{\sqrt{5} + 2 + \sqrt{5} - 2 + 2 \sqrt{5 - 4}}{\sqrt{5} + 1} = 2 \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = 2$ ${t a k e i t a s a}$ $a = \sqrt{2}$
	Answered by john santu last updated on 25/Jul/20

	$\frac{\sqrt{{(\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2})}^{2}}}{\sqrt{\sqrt{5} + 1}} =$ $\frac{\sqrt{{(\sqrt{\sqrt{5} + 2})}^{2} + 2 \sqrt{(\sqrt{5} + 2) (\sqrt{5} - 2)} + {(\sqrt{\sqrt{5} - 2})}^{2}}}{\sqrt{\sqrt{5} + 1}}$ $= \frac{\sqrt{\sqrt{5} + 2 + 2 \sqrt{5 - 4} + \sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}}$ $= \frac{\sqrt{2 \sqrt{5} + 2}}{\sqrt{\sqrt{5} + 1}} = \frac{\sqrt{2} . \sqrt{\sqrt{5} + 1}}{\sqrt{\sqrt{5} + 1}} = \sqrt{2}$ $(J S ⊛)$
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