Giải bài 15 trang 81 SGK giải tích nâng cao 12

${{\left( 0,{{5}^{\sqrt{2}}} \right)}^{\sqrt{8}}}={{\left( \dfrac{1}{2} \right)}^{4}}=\dfrac{1}{{{2}^{4}}}=\dfrac{1}{16} \\ {{2}^{2-3\sqrt{5}}}{{.8}^{\sqrt{5}}}={{2}^{2-3\sqrt{5}}}{{.2}^{3\sqrt{5}}}={{2}^{2-3\sqrt{5}+3\sqrt{5}}}={{2}^{2}}=4 \\ {{3}^{1+2\sqrt[3]{2}}}:{{9}^{\sqrt[3]{2}}}={{3}^{1+2\sqrt[3]{2}}}:{{3}^{2\sqrt[3]{2}}}={{3}^{1+2\sqrt[3]{2}-2\sqrt[3]{2}}}=3$

Ghi nhớ: ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\,\,\,\,\,\,\,\,\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},\,\,\,\,\,\,\,{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}$.