Giải bài 4 trang 76 SGK giải tích nâng cao 12

Hướng dẫn: Vận dụng các công thức: ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\,\,\,\,\,\,\,\,\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},\,\,\,\,\,\,\,\dfrac{1}{{{a}^{n}}}={{a}^{-n}},\,\,\,\,\,\,{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}$

Ta có 

$\begin{aligned} & a)\, {{81}^{-0,75}}+{{\left( \dfrac{1}{125} \right)}^{-\frac{1}{3}}}-{{\left( \dfrac{1}{32} \right)}^{-\frac{3}{5}}} \\ & ={{\left( {{3}^{4}} \right)}^{-\frac{3}{4}}}+{{\left( {{5}^{3}} \right)}^{\frac{1}{3}}}-{{\left( {{2}^{5}} \right)}^{\frac{3}{5}}} \\ & =\dfrac{1}{{{3}^{3}}}+5-{{2}^{3}} \\ & =\dfrac{1}{27}+5-8 \\ & =-\dfrac{80}{27} \\ \end{aligned}$

$\begin{align} & b)\, 0,{{001}^{-\frac{1}{3}}}-{{\left( -2 \right)}^{-2}}{{.64}^{\frac{2}{3}}}-{{8}^{-1\frac{1}{3}}}+{{\left( {{9}^{0}} \right)}^{2}} \\ & ={{\left( \dfrac{1}{1000} \right)}^{-\frac{1}{3}}}-\dfrac{1}{{{2}^{2}}}.{{\left( {{2}^{6}} \right)}^{\frac{2}{3}}}-{{\left( {{2}^{3}} \right)}^{-\frac{4}{3}}}+{{1}^{2}} \\ & ={{\left( {{10}^{3}} \right)}^{\frac{1}{3}}}-\dfrac{{{2}^{4}}}{{{2}^{2}}}-\dfrac{1}{{{\left( {{2}^{3}} \right)}^{\frac{4}{3}}}}+1 \\ & =10-{{2}^{4-2}}-\dfrac{1}{{{2}^{4}}}+1 \\ & =11-4-\dfrac{1}{16} \\ & =\dfrac{111}{16} \\ \end{align}$

$\begin{aligned} & c)\, {{27}^{\frac{2}{3}}}+{{\left( \dfrac{1}{16} \right)}^{-0,75}}-{{25}^{0,5}} \\ & ={{\left( {{3}^{3}} \right)}^{\frac{2}{3}}}+{{\left( {{2}^{4}} \right)}^ {\frac{3}{4}}}-{{\left( {{5}^{2}} \right)}^{\frac{1}{2}}} \\ & ={{3}^{2}}+{{2}^{3}}-5 \\ & =9+8-5 \\ & =12 \\ \end{aligned}$

$\begin{aligned} &d)\, {{\left( -0,5 \right)}^{-4}}- {{625}^{0,25}}-{{\left( 2\dfrac{1}{4} \right)}^{-1\frac{1}{2}}}+19{{\left( -3 \right)}^{-3}} \\ & ={{\left( \dfrac{1}{2} \right)}^{-4}}-{{\left( {{5}^{4}} \right)}^{\frac{1}{4}}}-{ {\left( \dfrac{9}{4} \right)}^{-\frac{3}{2}}}-\dfrac{19}{{{3}^{3}}} \\ & ={{2}^{4}}-5-{{\left( \dfrac{{{2}^{2}}}{{{3}^{2}}} \right)}^{\frac{3}{2}}}-\dfrac{19}{27} \\ & =16-\dfrac{{{2}^ {3}}}{{{3}^{3}}}-\dfrac{154}{27} \\ & =\dfrac{278}{27}-\dfrac{8}{27} \\ & =10 \end{aligned}$