Giải bài 18 trang 81 SGK giải tích nâng cao 12

$a)\,\sqrt[4]{{{x}^{2}}\sqrt[3]{x}}\,=\sqrt[4]{{{x}^{2+\frac{1}{3}}}}={{\left( {{x}^{\frac{7}{3}}} \right)}^{\frac{1}{4}}}={{x}^{\frac{7}{12}}} \\ b)\, \sqrt[5]{\dfrac{b}{a}\sqrt[3]{\dfrac{a}{b}}}={{\left( \dfrac{b}{a}{{\left( \dfrac{a}{b} \right)}^{\frac{1}{3}}} \right)}^{\frac{1}{5}}}={{\left( {{\left( \dfrac{b}{a} \right)}^{\frac{2}{3}}} \right)}^{\frac{1}{5}}}={{\left( \dfrac{b}{a} \right)}^{\frac{2}{15}}} \\ c)\, \sqrt[3]{\dfrac{2}{3}\sqrt[3]{\dfrac{2}{3}\sqrt{\dfrac{2}{3}}}}=\sqrt[3]{\dfrac{2}{3}\sqrt[3]{{{\left( \dfrac{2}{3} \right)}^{\frac{3}{2}}}}}=\sqrt[3]{{{\left( \dfrac{2}{3} \right)}^{\frac{3}{2}}}}={{\left( \dfrac{2}{3} \right)}^{\frac{1}{2}}}$

$\begin{align} d)\,&\sqrt{a\sqrt{a\sqrt{a\sqrt{a}}}}:{{a}^{\frac{11}{6}\,}}=\sqrt{a\sqrt{a\sqrt{{{a}^{\frac{3}{2}}}}}}:{{a}^{\frac{11}{6}\,}} \\ & =\sqrt{a\sqrt{{{a}^{\frac{7}{4}}}}}:{{a}^{\frac{11}{6}\,}}=\sqrt{{{a}^{\frac{15}{8}}}}:{{a}^{\frac{11}{6}\,}} \\ & ={{a}^{\frac{15}{16}}}:{{a}^{\frac{11}{6}\,}}={{a}^{\frac{1}{4}}} \\ \end{align}$

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