Giải bài 4 trang 56 – SGK môn Giải tích lớp 12
Cho \(a,\,b\) là những số thực dương. Rút gọn các biểu thức sau:
a) \(\dfrac{{{a}^{\frac{4}{3}}}\left( {{a}^{-\frac{1}{3}}} +{{a}^{\frac{2}{3}}} \right)}{{{a}^{\frac{1}{4}}} \left( {{a}^{\frac{3}{4}}}+{{a}^{-\frac{1}{4}}} \right)}\);
b) \(\dfrac{{{b}^{\frac{1}{5}}}\left( \sqrt[5]{{{b}^{4}}}-\sqrt[5]{{{b}^{-1}}} \right)} {{{b}^{\frac{2}{3}}}\left( \sqrt[3]{b}-\sqrt[3]{{{b}^{-2}}} \right)}\);
c) \(\dfrac{{{a}^{\frac{1}{3}}}{{b}^{-\frac{1}{3}}}-{{a}^{-\frac{1}{3}}} {{b}^{\frac{1}{3}}}}{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}\);
d) \(\dfrac{{{a}^{\frac{1}{3}}}\sqrt{b}+{{b}^{\frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\)
a) \( \dfrac{{{a}^{\frac{4}{3}}}\left( {{a}^{-\frac{1}{3}}} +{{a}^{\frac{2}{3}}} \right)}{{{a}^{\frac{1}{4}}} \left( {{a}^{\frac{3}{4}}}+{{a}^{-\frac{1}{4}}} \right)} =\dfrac{a+{{a}^{2}}}{a+1}=\dfrac{a\left( a+1 \right)}{a+1}=a\)
b) \(\dfrac{{{b}^{\frac{1}{5}}}\left( \sqrt[5]{{{b}^{4}}}-\sqrt[5]{{{b}^{-1}}} \right)} {{{b}^{\frac{2}{3}}}\left( \sqrt[3]{b}-\sqrt[3]{{{b}^{-2}}} \right)}= \dfrac{{{b}^{\frac{1}{5}}}\left( {{b}^{\frac{4}{5}}}-{{b}^{-\frac{1}{5}}} \right)}{{{b}^{\frac{2}{3}}} \left( {{b}^{\frac{1}{3}}}-{{b}^{\frac{-2}{3}}} \right)}=\dfrac{b-1}{b-1}=1\)
c) \(\dfrac{{{a}^{\frac{1}{3}}}{{b}^{-\frac{1}{3}}}-{{a}^{-\frac{1}{3}}} {{b}^{\frac{1}{3}}}}{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}=\dfrac{{{a}^{-\frac{1}{3}}} {{b}^{-\frac{1}{3}}}\left( {{a}^{\frac{2}{3}}}-{{b}^{\frac{2}{3}}} \right)} {{{a}^{\frac{2}{3}}}-{{b}^{\frac{2}{3}}}}={{a}^{-\frac{1}{3}}}{{b}^{-\frac{1}{3}}}=\dfrac{1}{\sqrt[3]{ab}}\)
d) \(\dfrac{{{a}^{\frac{1}{3}}}\sqrt{b}+{{b}^{\frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}} =\dfrac{{{a}^{\frac{1}{3}}}{{b}^{\frac{1}{3}}}\left( {{a}^{\frac{1}{6}}}+{{b}^{\frac{1}{6}}} \right)}{{{a}^{\frac{1}{6}}}+{{b}^{\frac{1}{6}}}} ={{a}^{\frac{1}{3}}}{{b}^{\frac{1}{3}}}=\sqrt[3]{ab} \)
Ghi nhớ: \(\sqrt[n]{{{a}^{m}}}={{a}^{\frac{m}{n}}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\,\,\,\,\,\,\,\,\,\,\,{{a}^{m}}:{{a}^{n}}={{a}^{m-n}}\).