Bài 4 (Trang 56 SGK Toán Giải Tích 12):

Cho a, b là những số thực dương. Rút gọn các biểu thức sau:

a) $\frac{{\mathrm{a}}^{{\displaystyle \frac{4}{3}}}\left({\mathrm{a}}^{-{\displaystyle \frac{1}{3}}}+{\mathrm{a}}^{{\displaystyle \frac{2}{3}}}\right)}{{\mathrm{a}}^{{\displaystyle \frac{1}{4}}}\left({\mathrm{a}}^{{\displaystyle \frac{3}{4}}}+{\mathrm{a}}^{-{\displaystyle \frac{1}{4}}}\right)}$;

b) $\frac{{\mathrm{b}}^{{\displaystyle \frac{1}{5}}}\left(\sqrt[5]{{\mathrm{b}}^4}-\sqrt[5]{{\mathrm{b}}^{-1}}\right)}{{\mathrm{b}}^{{\displaystyle \frac{2}{3}}}\left(\sqrt[3]{\mathrm{b}}-\sqrt[3]{{\mathrm{b}}^{-2}}\right)}$;

c) $\frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}{\mathrm{b}}^{-{\displaystyle \frac{1}{3}}}-{\mathrm{a}}^{-{\displaystyle \frac{1}{3}}}{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{{\mathrm{b}}^2}}$;

d) $\frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}\sqrt{\mathrm{b}}+{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}\sqrt{\mathrm{a}}}{\sqrt[6]{\mathrm{a}}+\sqrt[6]{\mathrm{b}}}$.

Hướng dẫn giải:

a) $\frac{{\mathrm{a}}^{{\displaystyle \frac{4}{3}}}\left({\mathrm{a}}^{-{\displaystyle \frac{1}{3}}}+{\mathrm{a}}^{{\displaystyle \frac{2}{3}}}\right)}{{\mathrm{a}}^{{\displaystyle \frac{1}{4}}}\left({\mathrm{a}}^{{\displaystyle \frac{3}{4}}}+{\mathrm{a}}^{-{\displaystyle \frac{1}{4}}}\right)} = \frac{\mathrm{a} + {\mathrm{a}}^2}{\mathrm{a} + 1} = \frac{\mathrm{a}(\mathrm{a}+1)}{\mathrm{a}+1} = \mathrm{a}$

b) $\frac{{\mathrm{b}}^{{\displaystyle \frac{1}{5}}}\left(\sqrt[5]{{\mathrm{b}}^4}-\sqrt[5]{{\mathrm{b}}^{-1}}\right)}{{\mathrm{b}}^{{\displaystyle \frac{2}{3}}}\left(\sqrt[3]{\mathrm{b}}-\sqrt[3]{{\mathrm{b}}^{-2}}\right)} = \frac{{\mathrm{b}}^{{\displaystyle \frac{1}{5}}}\left({\mathrm{b}}^{{\displaystyle \frac{4}{5}}}-{\mathrm{b}}^{-{\displaystyle \frac{1}{5}}}\right)}{{\mathrm{b}}^{{\displaystyle \frac{2}{3}}}\left({\mathrm{b}}^{{\displaystyle \frac{1}{3}}}-{\mathrm{b}}^{-{\displaystyle \frac{2}{3}}}\right)} = \frac{\mathrm{b}-1}{\mathrm{b}-1} = 1$

c) $\frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}{\mathrm{b}}^{-{\displaystyle \frac{1}{3}}}-{\mathrm{a}}^{-{\displaystyle \frac{1}{3}}}{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{{\mathrm{b}}^2}} = \frac{{\mathrm{a}}^{-{\displaystyle \frac{1}{3}}}.{\mathrm{b}}^{-{\displaystyle \frac{1}{3}}}\left({\mathrm{a}}^{{\displaystyle \frac{2}{3}}}-{\mathrm{b}}^{{\displaystyle \frac{2}{3}}}\right)}{{\mathrm{a}}^{{\displaystyle \frac{2}{3}}}-{\mathrm{b}}^{{\displaystyle \frac{2}{3}}}} = {\mathrm{a}}^{-\frac{1}{3}}.{\mathrm{b}}^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{\mathrm{ab}}}$;

d) $\frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}\sqrt{\mathrm{b}}+{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}\sqrt{\mathrm{a}}}{\sqrt[6]{\mathrm{a}}+\sqrt[6]{\mathrm{b}}} = \frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}.{\mathrm{b}}^{{\displaystyle \frac{1}{2}}}+{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}.{\mathrm{a}}^{{\displaystyle \frac{1}{2}}}}{{\mathrm{a}}^{{\displaystyle \frac{1}{6}}}+{\mathrm{b}}^{{\displaystyle \frac{1}{6}}}} = \frac{{\mathrm{a}}^{{\displaystyle \frac{1}{3}}}.{\mathrm{b}}^{{\displaystyle \frac{1}{3}}}\left({\mathrm{b}}^{{\displaystyle \frac{1}{6}}}+{\mathrm{a}}^{{\displaystyle \frac{1}{6}}}\right)}{{\mathrm{a}}^{\frac{1}{6}}+{\mathrm{b}}^{\frac{1}{6}}} = {\mathrm{a}}^{\frac{1}{3}}.{\mathrm{b}}^{\frac{1}{3}} = \sqrt[3]{\mathrm{ab}}$