Find the value of x+3 x-3 + x+2 x-2 , if x= 2 6 3+2 . - Vidyadip

Question

Find the value of
$\frac{x+\sqrt{3}}{x-\sqrt{3}}+\frac{x+\sqrt{2}}{x-\sqrt{2}}$, if $\quad x=\frac{2 \sqrt{6}}{\sqrt{3}+\sqrt{2}}$.

✓

Answer

$\frac{2 \sqrt{6}}{\sqrt{3}+\sqrt{2}}$
or
$ x =\frac{2 \times \sqrt{3} \times \sqrt{2}}{\sqrt{3}+\sqrt{2}} $
$ \Rightarrow \frac{x}{\sqrt{3}}=\frac{2 \sqrt{2}}{\sqrt{3}+\sqrt{2}}$
Applying Componendo and Dividendo
$\frac{x+\sqrt{3}}{x-\sqrt{3}}=\frac{2 \sqrt{2}+\sqrt{3}+\sqrt{2}}{2 \sqrt{2}-\sqrt{3}-\sqrt{2}}$
$ =\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
$\frac{x+\sqrt{3}}{x-\sqrt{3}}=\frac{3 \sqrt{2}+\sqrt{3}}{-(\sqrt{3}-\sqrt{2})}$...(i)
Also $\frac{x}{\sqrt{2}}=\frac{2 \sqrt{3}}{\sqrt{3}+\sqrt{2}}$
Applying Componendo and Dividendo
$\frac{x+\sqrt{2}}{x+\sqrt{2}}=\frac{2 \sqrt{3}+\sqrt{3}+\sqrt{2}}{2 \sqrt{3}-\sqrt{3}-\sqrt{2}}$
$\frac{x+\sqrt{2}}{x-\sqrt{2}}=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ ..(ii)
$\frac{x+\sqrt{3}}{x-\sqrt{3}}+\frac{x+\sqrt{2}}{x-\sqrt{2}}=\frac{-3 \sqrt{2}-\sqrt{3}+3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} $
$=\frac{-2 \sqrt{2}+2 \sqrt{3}}{\sqrt{3}-\sqrt{2}}$
$ =\frac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})}$
$ =2 .$

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