Answer
$\frac{\sqrt2-3}{\sqrt2-3}$
$\frac{2(\sqrt2-3)}{2(\sqrt2-3)}$
$\frac{10(\sqrt2-3)}{10(\sqrt2-3)}$
All of these will be the same since any number divided by itself is 1.
Work Step by Step
$\frac{\sqrt2-3}{\sqrt2-3}*\frac1{\sqrt2+3}$
$\frac{\sqrt2-3}{2-9}$
$\frac{\sqrt2-3}{-7}$
$\frac{10(\sqrt2-3)}{10(\sqrt2-3)}*\frac1{\sqrt2+3}$
$\frac{10(\sqrt2-3)}{10*({2-9})}$
$\frac{\sqrt2-3}{{2-9}}$
$\frac{\sqrt2-3}{{-7}}$
