Question

if x = 3-2
$\sqrt{2}$
find the value of
$\sqrt{x} + 1 \div \sqrt{x}$




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Answer 1

Answer:

$x = 3 - 2 \sqrt{2} \\ \\ \\ \\ \sqrt{x} + \frac{1}{ \sqrt{x} } = > \sqrt{3 - 2 \sqrt{2} } + \frac{1}{ \sqrt{3 - 2 \sqrt{2} } } \\ \\ = \\ \\ \\ = > \frac{3 - 2 \sqrt{2} + 1}{ \sqrt{3 - 2 \sqrt{2} } } \\ \\ = > \frac{4 - 2 \sqrt{2} }{ \sqrt{3} - 2 \sqrt{2} }$

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