Question

If x=9 +4√5 ; find Set √x+1÷√x
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Answer 1

Answer:

tittitcv rryufi cicifig

$\sqrt[6115 {46163y4) { \frac{ \frac{ {y \sqrt{ \frac{1222}{?} } }^{?} }{?} }{?} }^{2} }^{2} ]{?}$

Answer 2

$\bf \huge \underbrace{Answer} \\ \\ x = 9 + 4 \sqrt{5} \\ \\ \sqrt{x} = \sqrt{9 + 4 \sqrt{5} } \\ \\ \sqrt{x} = \sqrt{5 + 4 + 4 \sqrt{5} } \\ \\ \sqrt{x} = \sqrt{ {( \sqrt{5} )}^{2} + {2}^{2} + 2 \times \sqrt{5} \times 2 } \\ \\ \sqrt{x} = \sqrt{5} + 2 \\ \\ and \: \: \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + 2} \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + 2} \times \frac{ \sqrt{5} - 2}{ \sqrt{5} - 2 } \\ \\ \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - 2}{ {( \sqrt{5} })^{2} - {(2)}^{2} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - 2}{5 - 4} \\ \\ \frac{1}{ \sqrt{x} } = \sqrt{5} - 2 \\ \\ now \\ \\ \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{5} + 2 + \sqrt{5} - 2 = 2 \sqrt{5}$