308. a) $\frac{1}{\sqrt{2}}$;
b) $\frac{3}{\sqrt{40}}$;
c) $\frac{2}{\sqrt[5]{3^{2}}}$;
d) $\frac{2}{3+\sqrt{5}}$.
Solution. a) to get rid of the root, multiply the numerator and denominator by $\sqrt{2}$ :

$$
\frac{1}{\sqrt{2}}=\frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{2}
$$

b) Multiplying the numerator and denominator by $\sqrt{10}$ and considering that $\sqrt{40}=$ $=\sqrt{4 \cdot 10}=2 \sqrt{10}$, we get

$$
\frac{3}{\sqrt{40}}=\frac{3 \sqrt{10}}{\sqrt{40} \cdot \sqrt{10}}=\frac{3 \sqrt{10}}{2 \sqrt{10} \cdot \sqrt{10}}=\frac{3 \sqrt{10}}{20}
$$

c) Multiply the numerator and denominator by $\sqrt[5]{3^{3}}$ :

$$
\frac{2}{\sqrt[5]{3^{2}}}=\frac{2 \sqrt[5]{3^{3}}}{\sqrt[5]{3^{2}} \cdot \sqrt[5]{3^{3}}}=\frac{2 \sqrt[5]{3^{3}}}{\sqrt[5]{3^{5}}}=\frac{2}{3} \sqrt[5]{27}
$$

d) Multiplying the numerator and denominator by the conjugate binomial $3-\sqrt{5}$, we get

$$
\frac{2}{3+\sqrt{5}}=\frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{2(3-\sqrt{5})}{9-5}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}
$$