Giải bài 5 trang 76 SGK giải tích nâng cao 12

$\begin{align} a)\, \dfrac{{{\left( \sqrt[4]{{{a}^{3}}{{b}^{2}}} \right)}^{4}}}{\sqrt[3]{\sqrt{{{a}^{12}}{{b}^{6}}}}}&=\dfrac{{{\left( {{a}^{3}}{{b}^{2}} \right)}^{4}}{{^{.}}^{\frac{1}{4}}}}{{{\left( {{a}^{12}}{{b}^{6}} \right)}^{\frac{1}{2}.\frac{1}{3}}}} \\ & =\dfrac{{{a}^{3}}{{b}^{2}}}{{{\left( {{a}^{12}}{{b}^{6}} \right)}^{\frac{1}{6}}}} \\ & =\dfrac{{{a}^{3}}{{b}^{2}}}{{{a}^{2}}b}=ab \\ \end{align}$

$\begin{align} &b) \dfrac{{{a}^{\frac{1}{3}}}-{{a}^{\frac{7}{3}}}}{{{a}^{\frac{1}{3}}}-{{a}^{\frac{4}{3}}}}-\dfrac{{{a}^{-\frac{1}{3}}}-{{a}^{\frac{5}{3}}}}{{{a}^{\frac{2}{3}}}+{{a}^{-\frac{1}{3}}}} \\ & =\dfrac{{{a}^{\frac{1}{3}}}\left( 1-{{a}^{2}} \right)}{{{a}^{\frac{1}{3}}}\left( 1-a \right)}-\frac{{{a}^{-\frac{1}{3}}}\left( 1-{{a}^{2}} \right)}{{{a}^{-\frac{1}{3}}}\left( a+1 \right)} \\ & =1+a-\left( 1-a \right) \\ & =2a \\ \end{align}$