Question

if x = 9 + 4√5, find √x - 1 / √x

Answer 1

x=5+4+2(2)√5
x=(√5)^2+(2)^2+2(2)√5
x=(√5+2)^2
√x=√5+2
1/√x=1/√5+2
          =√5-2/5-4
              =√5-2
√x - 1 / √x=√5+2-(√5-2)
                  =√5+2-√5+2
                    =4

Answer 2

Hello yuvraj!

#ur Ans
__________

Given that
X = 9+4√5
so

$\frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \\ \\ now \: rationalising \: denometer \\ \\ \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ = > \frac{1}{x} = \frac{9 - 4 \sqrt{5} }{ {9}^{2} - (4 \sqrt{5} {)}^{2} } \\ \\ = > \frac{1}{x} = \frac{9 - 4 \sqrt{5} }{81 - 80} \\ \\ = > \frac{1}{x} = 9 - 4 \sqrt{5} \\ \\ now \\ \\ ( \sqrt{x} + \frac{1}{ \sqrt{x} } {)}^{2} = x + \frac{1}{x} + 2 \\ \\ = > ( \sqrt{x} + \frac{1}{ \sqrt{x} {}^{} } {)}^{2} = 9 + 4 \sqrt{5 } + 9 - 4 \sqrt{5} + 2 \\ \\ = > ( \sqrt{x} + \frac{1}{ \sqrt{x} } {)}^{2} = 20 \\ \\ = > \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{20} \\ \\ = > \sqrt{x} + \frac{1}{ \sqrt{x} } = 2 \sqrt{5} \: ans$

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