题目内容
17.化简:(1)${a}^{\frac{1}{3}}$•${a}^{\frac{3}{4}}$•${a}^{\frac{7}{12}}$;
(2)${a}^{\frac{3}{2}}$•${a}^{\frac{3}{4}}$÷${a}^{\frac{5}{6}}$;
(3)3${a}^{\frac{3}{2}}$•(-a${\;}^{\frac{3}{4}}$)÷9$\sqrt{a}$;
(4)$\frac{{a}^{2}}{\sqrt{a}•\root{3}{{a}^{2}}}$;
(5)${(\frac{{8a}^{-3}}{2{7b}^{6}})}^{-\frac{1}{3}}$;
(6)2x${\;}^{\frac{1}{3}}$($\frac{1}{2}$${x}^{\frac{1}{3}}$-2x${\;}^{\frac{2}{3}}$);
(7)(a${\;}^{\frac{8}{5}}$b${\;}^{-\frac{6}{5}}$)${\;}^{-\frac{1}{2}}$•$\root{5}{{a}^{4}}$÷$\root{5}{{b}^{3}}$(a≠0,b≠0);
(8)(2a${\;}^{\frac{2}{3}}$b${\;}^{\frac{1}{2}}$)(-6a${\;}^{\frac{1}{2}}$b${\;}^{\frac{1}{3}}$)÷(-3a${\;}^{\frac{1}{6}}$b${\;}^{\frac{5}{6}}$)
分析 先把根式化为分数指数幂的形式,再利用幂的运算法则进行计算即可.
解答 解:(1)原式=${a}^{\frac{1}{3}+\frac{3}{4}+\frac{7}{12}}$=${a}^{\frac{5}{3}}$;
(2)原式=${a}^{\frac{3}{2}+\frac{3}{4}-\frac{5}{6}}$=${a}^{\frac{17}{12}}$;
(3)原式=3×(-1)÷9•${a}^{\frac{3}{2}+\frac{3}{4}-\frac{1}{2}}$=-$\frac{1}{3}$${a}^{\frac{7}{4}}$;
(4)原式=$\frac{{a}^{2}}{{a}^{\frac{1}{2}}{•a}^{\frac{2}{3}}}$=${a}^{2-(\frac{1}{2}+\frac{2}{3})}$=${a}^{\frac{5}{6}}$;
(5)原式=$\frac{{8}^{-\frac{1}{3}}{•a}^{-3×(-\frac{1}{3})}}{{27}^{-\frac{1}{3}}{•b}^{6×(-\frac{1}{3})}}$=$\frac{{2}^{-1}•a}{{3}^{-1}{•b}^{-2}}$=$\frac{3}{2}$ab2;
(6)原式=(2×$\frac{1}{2}$)•${x}^{\frac{1}{3}}$•${x}^{\frac{1}{3}}$-(2×2)•${x}^{\frac{1}{3}}$•${x}^{\frac{2}{3}}$=${x}^{\frac{2}{3}}$-4x;
(7)原式=${a}^{\frac{8}{5}×(-\frac{1}{2})}$•${b}^{-\frac{6}{5}×(-\frac{1}{2})}$•${a}^{\frac{4}{5}}$÷${b}^{\frac{3}{5}}$=${a}^{-\frac{4}{5}+\frac{4}{5}}$•${b}^{\frac{3}{5}-\frac{3}{5}}$=a0•b0=1;
(8)原式=[2×(-6)÷(-3)]•${a}^{\frac{2}{3}+\frac{1}{2}-\frac{1}{6}}$•${b}^{\frac{1}{2}+\frac{1}{3}-\frac{5}{6}}$=4a.
点评 本题考查了根式化为分数指数幂的运算问题,也考查了幂的运算法则的应用问题,是基础题目.
A. | $\frac{25}{6}$ | B. | $\frac{25}{3}$ | C. | $\frac{50}{3}$ | D. | 25 |