## Precalculus (6th Edition) Blitzer

$\frac{3}{4}$
Recall that any number raised to the $0^\text{th}$ power is 1, and that $x^{m/n}=\sqrt[n] {x^m}$. We use these rules and the basic laws of exponents to simplify: $$\left(\frac{2}{7}\right)^0-32^{-2/5}$$$$1-\frac{1}{32^{2/5}}$$$$1-\frac{1}{\sqrt[5] {32^2}}$$ $$1-\frac{1}{4}$$$$\frac{3}{4}$$