Question

tell how to solve this sum !?.

Answer 1

Heya !!

Refer to the attachment

Hope it helps !!

Answer 2

Finding the value of $x+\frac{1}{x}$  as follows:

Step-by-step explanation:

$\ Given \ that :$

$x=\sqrt{2}+1$

$\ and \ find \ x+\frac{1}{x} \ = \ ?$  

$\ let, \ x+\frac{1}{x} \ ....... (1)$

$\ put \ the \ value \ of \ x \ in \ equation (1)\\\\(\sqrt{2} +1) +\frac{1}{\sqrt{2} +1 }\\\\\ taking \ L.C.M \ and \ add \ value\\\\\frac{(\sqrt{2}+1)^2+ 1}\sqrt{2}+1} \\\\\\ formula: \ (a+b)^2= a^2+ b^2+ 2\cdot a \cdot b\\\\\ (\sqrt{2})^2 +(1)^2+2 \cdot \sqrt{2} \cdot 1\\\\2+1+2\sqrt{2} \ = \ 3+2\sqrt{2}$

$\frac{\ 3+2\sqrt{2}+1}{\sqrt{2} +1}$

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

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

$\frac{\ 2(2+\sqrt{2)}}{\sqrt{2} +1} \times \frac{\sqrt{2} -1}{\sqrt{2} -1}$

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

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

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

${2\sqrt{2}}$

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