Precalculus (6th Edition) Blitzer

$\dfrac{\sqrt{10}}{5}$
RECALL: (i) For any nonnegative real number $a$, $\sqrt{a} \cdot \sqrt{a} = a$. (ii) For any nonnegative real numbers a and b, $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$. Rationalize the denominator by multiplying $\sqrt{5}$ to the numerator and the denominator. Then, use the rules above to obtain $\dfrac{\sqrt{2}}{\sqrt{5}} \cdot \dfrac{\sqrt{5}}{\sqrt{5}} \\=\dfrac{\sqrt{10}}{5}.$ Simplify by canceling the common factor to obtain $\require{cancel} \dfrac{\cancel{2}\sqrt{10}}{\cancel{2}(5)} \\=\dfrac{\sqrt{10}}{5}.$