【学以致用1】(1)(多选)下列哪些函数是幂函数()
A. $y = 2^{x}$
B. $y = x^{-2}$
C. $y = (2x)^{3}$
D. $y = \frac{1}{x}$
(2)已知点$(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{4})在幂函数y = f(x)$的图象上,则$y = f(x)$的解析式为______.

(1)BD (2)$y = f(x) = x^3$ (1)形如$y = x^a$的函数为幂函数，故AC错误，B正确；$y = \frac{1}{x} = x^{-1}$为幂函数，故D正确。
(2)设幂函数$y = f(x) = x^a$。
代入点$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4})$，得$(\frac{\sqrt{2}}{2})^a = \frac{\sqrt{2}}{4}$，解得$a = 3$，所以幂函数$y = f(x) = x^3$。

【学以致用2】已知幂函数$f(x)过点(2,\sqrt{2})$,若$f(10 - 2a) \lt f(a + 1)$,则实数$a$的取值范围是______.

(3,5]　设幂函数的解析式为$f(x) = x^a$。根据题意，得$2^a = \sqrt{2} = 2^{\frac{1}{2}}$，所以$a = \frac{1}{2}$，所以$f(x) = \sqrt{x}$，故$f(x)$在$[0, +\infty)$上单调递增。因为$f(10 - 2a) ＜ f(a + 1)$，所以$\begin{cases}a + 1 \geq 0, \\ 10 - 2a \geq 0, \\ 10 - 2a ＜ a + 1\end{cases}$，解得$\begin{cases}a \geq -1, \\ a \leq 5, \\ a ＞ 3\end{cases}$，所以$3 ＜ a \leq 5$。