Giải bài 2.2 trang 99 - SBT Giải tích lớp 12
a) \(27^{\frac 2 3}-(-2)^{-2}+\left(3\dfrac 3 8\right)^{-\frac 1 3}\)
b) \((-0,5)^{-4}-625^{0,25}-\left(2\dfrac 1 4\right)^{-1\frac 1 2}\)
Hướng dẫn:
Áp dụng:
\(a^{-m}=\dfrac 1 {a^m}\,\,\,(a\ne 0;\,m>0)\\ {(a^m)}^{n}=a^{m.n}\)
a)
\(27^{\frac 2 3}-(-2)^{-2}+\left(3\dfrac 3 8\right)^{-\frac 1 3}\\ \begin{align} & ={{\left( {{3}^{3}} \right)}^{\frac{2}{3}}}-\frac{1}{{{\left( -2 \right)}^{2}}}+{{\left( \frac{27}{8} \right)}^{-\frac{1}{3}}} \\ & ={{3}^{3.\frac{2}{3}}}-\frac{1}{4}+{{\left[ {{\left( \frac{3}{2} \right)}^{3}} \right]}^{-\frac{1}{3}}}={{3}^{2}}-\frac{1}{4}+{{\left( \frac{3}{2} \right)}^{-1}} \\ & =9-\frac{1}{4}+\frac{2}{3}=\frac{113}{12} \\ \end{align} \)
b)
\((-0,5)^{-4}-625^{0,25}-\left(2\dfrac 1 4\right)^{-1\frac 1 2}\\ \begin{align} & ={{\left( \frac{-1}{2} \right)}^{-4}}-{{\left( {{5}^{4}} \right)}^{\frac{1}{4}}}-{{\left( \frac{9}{4} \right)}^{-\frac{3}{2}}} \\ & ={{\left( -2 \right)}^{4}}-{{5}^{4.\frac{1}{4}}}-{{\left[ {{\left( \frac{3}{2} \right)}^{2}} \right]}^{-\frac{3}{2}}} \\ & =16-5-{{\left( \frac{3}{2} \right)}^{-3}} \\ & =11-{{\left( \frac{2}{3} \right)}^{3}}=11-\frac{8}{27}=\frac{289}{27} \\ \end{align} \)