Question

If $x=2\sqrt{3}+2\sqrt{2}$, then find the value of : $\frac{1}{x}$.

Answer 1

Given that,

$x=2\sqrt{3}+2\sqrt{2}=2(\sqrt{3}+\sqrt{2})$

$\frac{1}{x}=\frac{1}{2(\sqrt{3}+\sqrt{2})}=\frac{1}{2(\sqrt{3}+\sqrt{2})}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{2\left[\right(\sqrt{3}{)}^2-\left(\sqrt{2}{)}^2\right]}$

$=\frac{\sqrt{3}-\sqrt{2}}{2(3-2)}$

$=\frac{\sqrt{3}-\sqrt{2}}{2\left(1\right)}$

$=\frac{\sqrt{3}-\sqrt{2}}{2}$

$\therefore \frac{1}{x}=\frac{\sqrt{3}-\sqrt{2}}{2}$