## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{x^{2}}{7y^{3/2}}$
Note: assuming all variables are positive, we do not need the absolute value brackets. For positive $a,$ $(a^{2})^{1/2}=\sqrt{a^{2}}=|a|$, the square root is $a.$ $(49x^{-2}y^{4})^{-1/2}(xy^{1/2})=\qquad$... apply $(ab)^{n}=a^{n}b^{n}$ $=(49)^{-1/2}(x^{-2})^{-1/2}(y^{4})^{-1/2}(xy^{1/2})\qquad$... apply $(a^{m})^{n}=a^{mn}$ $=(49)^{-1/2}(x^{-2\cdot(-1/2)})y^{4(-1/2)}(xy^{1/2})\qquad$... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $=\displaystyle \frac{1}{49^{1/2}}x^{1}y^{-2}\cdot x^{1}y^{1/2}\qquad$... apply $a^{m}a^{n}=a^{m+n}$ $=\displaystyle \frac{1}{7}x^{1+1}y^{-2+(1/2)}$ $=\displaystyle \frac{1}{7}x^{2}y^{-3/2}\qquad$... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $=\displaystyle \frac{x^{2}}{7y^{3/2}}$