Answer
$\dfrac{\sqrt{r}+\sqrt{2}}{5}=\dfrac{r-2}{5\sqrt{r}-5\sqrt{2}}$
Work Step by Step
$\dfrac{\sqrt{r}+\sqrt{2}}{5}$
Multiply both numerator and denominator by $\sqrt{r}-\sqrt{2}$ and simplify:
$\dfrac{\sqrt{r}+\sqrt{2}}{5}=\Big(\dfrac{\sqrt{r}+\sqrt{2}}{5}\Big)\Big(\dfrac{\sqrt{r}-\sqrt{2}}{\sqrt{r}-\sqrt{2}}\Big)=\dfrac{(\sqrt{r})^{2}-(\sqrt{2})^{2}}{5(\sqrt{r}-\sqrt{2})}=...$
$...=\dfrac{r-2}{5\sqrt{r}-5\sqrt{2}}$
