Question

x=
$\sqrt{2} + 1 \\ \\ \\ x + \frac{1}{x}$
​

Answer 1

$( \sqrt{2} + 1) + 1 \div (\sqrt{2} + 1)$

$(2 + 1 + 2 \sqrt{2} + 1) \div \sqrt{2} + 1$

$(4 + 2 \sqrt{2} )\div (1 + \sqrt{2} )$

Answer 2

Answer:

2√2

Step-by-step explanation:

by solving equation

$x+\frac{1}{x} = \frac{x^{2}+1}{x}$

(by L.CM method )

substituting the value of x in equation

$\frac{x^{2}+1}{x}= \frac{({\sqrt{2}+1)}^2+1 }{\sqrt{2}+1 }$

$=\frac{{(\sqrt{2})}^2+{(1)}^2+2(\sqrt{2})(1) }{\sqrt{2}+1 }$

(by the formula of (a+b)²

$\frac{(\sqrt{2})^{2}+{1}^2 +2\sqrt{2} +1}{\sqrt{2}+1 } =\frac{2+1+2\sqrt{2}+1 }{\sqrt{2}+1 }$

=

$\frac{2+1+2\sqrt{2}+1 }{\sqrt{2}+1 } =\frac{4+2\sqrt{2} }{\sqrt{2}+1 }$

=

$\frac{4+2\sqrt{2} }{\sqrt{2}+1 } =\frac{2\sqrt{2}(\sqrt{2}+1 )}{\sqrt{2}+1 }$

=

$\frac{2\sqrt{2}(\sqrt{2}+1 )}{\sqrt{2}+1 } =2\sqrt{2}$

hope you will like my answer