Giải bài 1 trang 55 – SGK môn Giải tích lớp 12
Tính:
a) \({{9}^{\frac{2}{5}}}{{.27}^{\frac{2}{5}}}\) ;
b) \({{144}^{\frac{3}{4}}}:{{9}^{\frac{3}{4}}}\);
c) \({{\left( \dfrac{1}{16} \right)}^{-0,75}}+0,{{25}^{-\frac{5}{2}}}\);
d) \({{\left( 0,04 \right)}^{-1,5}}-{{\left( 0,125 \right)}^{-\frac{2}{3}}}\)
Hướng dẫn: Vận dụng các công thức:
\({{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\,\,\,\,\,\,\,\, \dfrac{{{a}^{m}}}{{{b}^{m}}}=\left(\dfrac{{{a}}}{{{a}}}\right)^m, \,\,\,\,\,\,\,\dfrac{1}{{{a}^{n}}}={{a}^{-n}},\,\,\,\,\,\,{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}\)
a) \({{9}^{\frac{2}{5}}}{{.27}^{\frac{2}{5}}}={{\left( 9.27 \right)}^{\frac{2}{5}}}={{\left( {{3}^{5}} \right)}^{\frac{2}{5}}}={{3}^{2}}=9 \)
b) \({{144}^{\frac{3}{4}}}:{{9}^{\frac{3}{4}}}= {{\left( \dfrac{144}{9} \right)}^{\frac{3}{4}}}= {{16}^{\frac{3}{4}}}={{\left( {{2}^{4}} \right)}^{\frac{3}{4}}}={{2}^{3}}=8\)
c) \({{\left( \dfrac{1}{16} \right)}^{-0,75}}+0,{{25}^{-\frac{5}{2}}}={{\left( {{2}^{-4}} \right)}^{\frac{-3}{4}}} +{{\left( {{2}^{-2}} \right)}^{\frac{-5}{2}}}={{2}^{3}}+{{2}^{5}}=8+32=40\)
d) \({{\left( 0,04 \right)}^{-1,5}}-{{\left( 0,125 \right)}^{-\frac{2}{3}}} ={{\left( \dfrac{1}{25} \right)}^{\frac{-3}{2}}}-{{\left( \frac{1}{8} \right)}^{-\frac{2}{3}}} ={{\left( {{5}^{-2}} \right)}^{\frac{-3}{2}}}-{{\left( {{2}^{-3}} \right)}^{-\frac{2}{3}}}={{5}^{3}}-{{2}^{2}}=125-4=121\)