题目内容
6.计算下列各式的值:(1)${0.027^{-\frac{1}{3}}}-{(-\frac{1}{7})^{-2}}+{81^{\frac{3}{4}}}-{3^{-1}}+{(\sqrt{2}-1)^0}$
(2)log3$\frac{{\root{4}{27}}}{3}+lg25+lg4+{7^{{{log}_7}2}}$.
分析 (1)化负指数为正指数,化0指数幂为1,再由有理指数幂的运算性质得答案;
(2)化根式为分数指数幂,再由对数的运算性质化简得答案.
解答 解:(1)${0.027^{-\frac{1}{3}}}-{(-\frac{1}{7})^{-2}}+{81^{\frac{3}{4}}}-{3^{-1}}+{(\sqrt{2}-1)^0}$=$[(\frac{3}{10})^{3}]^{-\frac{1}{3}}-49+({3}^{4})^{\frac{3}{4}}-\frac{1}{3}+1$
=$\frac{10}{3}-49+27-\frac{1}{3}+1=-18$;
(2)log3$\frac{{\root{4}{27}}}{3}+lg25+lg4+{7^{{{log}_7}2}}$=$lo{g}_{3}\root{4}{27}-lo{g}_{3}3+lg(25×4)+2$
=$\frac{3}{4}-1+2+2$=$\frac{15}{4}$.
点评 本题考查了有理指数幂的化简求值,考查了对数的运算性质,是基础题.
练习册系列答案
相关题目
14.已知函数f(x)=$\left\{\begin{array}{l}{kx-1,x≤0}\\{{2}^{-x}-1,x>0}\end{array}\right.$,(k<0),当方程f[f(x)]=-$\frac{1}{2}$恰有三个实数根时,实数k的取值范围为( )
| A. | (-$\frac{1}{2}$,0) | B. | [-$\frac{1}{2}$,0) | C. | (-∞,-$\frac{1}{2}$] | D. | (-∞,-$\frac{1}{2}$) |
1.已知函数f(x)=2sin(ωx+$\frac{π}{6}$)-1(ω>0)在x∈[0,π]恰有3个零点,则实数ω取值范围为( )
| A. | [$\frac{5}{3}$,$\frac{8}{3}$] | B. | [2,$\frac{8}{3}$) | C. | [$\frac{5}{3}$,2] | D. | [$\frac{5}{3}$,2) |
18.若x,y满足$\left\{\begin{array}{l}x+y≤4\\ x-2y≥0\\ x+2y≥4\end{array}$则z=$\frac{y-4}{x}$的取值范围是( )
| A. | $(-∞,-\frac{3}{2}]∪[-1,+∞)$ | B. | $(-∞,-\frac{5}{2}]∪[-1,+∞)$ | C. | $[-\frac{5}{2},-\frac{3}{2}]$ | D. | $[-\frac{3}{2},-1]$ |
15.设函数f(x)=$\left\{\begin{array}{l}{3x-b(x<1)}\\{{3}^{x}(x≥1)}\end{array}\right.$,若$f(f(\frac{1}{2}))=9$,则实数b的值为( )
| A. | $-\frac{3}{2}$ | B. | $-\frac{9}{8}$ | C. | $-\frac{3}{4}$ | D. | $-\frac{1}{2}$ |