Giải bài 16 trang 81 SGK giải tích nâng cao 12

$*)\,\dfrac{{{\left( {{a}^{\sqrt{3}-1}} \right)}^{\sqrt{3}+1}}}{{{a}^{\sqrt{5}-3}}.{{a}^{4-\sqrt{5}}}}=\dfrac{{{a}^{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}}}{{{a}^{\sqrt{5}-3+4-\sqrt{5}}}}=\dfrac{{{a}^{2}}}{a}=a \\ *)\, {{a}^{\sqrt{2}}}.{{\left( \dfrac{1}{a} \right)}^{\sqrt{2}-1}}={{a}^{\sqrt{2}}}.{{a}^{1-\sqrt{2}}}=a$

Ghi nhớ: ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\,\,\,\,\,\,\,\,\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},\,\,\,\,\,\,\,\dfrac{1}{{{a}^{n}}}={{a}^{-n}},\,\,\,\,\,\,{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}$.