题目内容
18.化简(式中字母均为正数):(1)a${\;}^{\frac{1}{3}}$a${\;}^{\frac{3}{4}}$a${\;}^{\frac{7}{12}}$;
(2)(x${\;}^{\sqrt{3}}$y${\;}^{-\frac{\sqrt{3}}{4}}$)${\;}^{\frac{1}{\sqrt{3}}}$;
(3)4x${\;}^{\frac{1}{\sqrt{2}}}$(-3x${\;}^{-\frac{1}{\sqrt{2}}}$y2);
(4)($\frac{16{s}^{2}{t}^{-6}}{25{r}^{4}}$)${\;}^{-\frac{3}{2}}$.
分析 (1)(2)(3)(4)利用指数幂的运算性质即可得出.
解答 解:(1)原式=${a}^{\frac{1}{3}+\frac{3}{4}+\frac{7}{12}}$=${a}^{\frac{5}{3}}$.
(2)原式=${x}^{\sqrt{3}×\frac{1}{\sqrt{3}}}$${y}^{-\frac{\sqrt{3}}{4}×\frac{1}{\sqrt{3}}}$=$x{y}^{-\frac{1}{4}}$.
(3)原式=$-12{x}^{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}$y2=-12y2.
(4)原式=$(\frac{4}{5})^{2×(-\frac{3}{2})}$${s}^{2×(-\frac{3}{2})}$${t}^{-6×(-\frac{3}{2})}$${r}^{-4×(-\frac{3}{2})}$
=$\frac{125}{64}$s-3t9r6.
点评 本题考查了指数幂的运算性质,考查了推理能力与计算能力,属于基础题.
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