Giải bài 19 trang 82 SGK giải tích nâng cao 12

$\begin{align} a)\,{{a}^{-2\sqrt{2}}}{{\left( \dfrac{1}{{{a}^{-\sqrt{2}-1}}} \right)}^{\sqrt{2}+1}}&={{a}^{-2\sqrt{2}}}.{{\left( {{a}^{1+\sqrt{2}}} \right)}^{\sqrt{2}+1}} \\ & ={{a}^{-2\sqrt{2}}}.{{a}^{3+2\sqrt{2}}} \\ & ={{a}^{3}} \\ \end{align}$

$\begin{align} b)\,{{\left( \dfrac{{{a}^{\sqrt{3}}}}{{{b}^{\sqrt{3}-1}}} \right)}^{\sqrt{3}+1}}.\dfrac{{{a}^{-1-\sqrt{3}}}}{{{b}^{-2}}} &=\dfrac{{{a}^{3+\sqrt{3}}}}{{{b}^{2}}}.\dfrac{{{b}^{2}}}{{{a}^{1+\sqrt{3}}}} \\ & =\dfrac{{ {{a}^{3+\sqrt{3}}} }}{{{a}^{1+\sqrt{3}}}} \\ & ={{ {{a}^{3+\sqrt{3}-(1+\sqrt{3})}} }} \\ & ={{a}^{2}} \\ \end{align}$

$\begin{align} & c)\,\dfrac{{{a}^{2\sqrt{2}}}-{{b}^{2\sqrt{3}}}}{{{\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}^{2}}}+1 \\ & =\dfrac{{{a}^{2\sqrt{2}}}-{{b}^{2\sqrt{3}}}+{{\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}^{2}}}{{{\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}^{2}}} \\ & =\dfrac{2{{a}^{2\sqrt{2}}}-2{{a}^{\sqrt{2}}}{{b}^{\sqrt{3}}}}{{{\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}^{2}}} \\ & =\dfrac{2{{a}^{\sqrt{2}}}\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}{{{\left( {{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}} \right)}^{2}}} \\ & =\dfrac{2{{a}^{\sqrt{2}}}}{{{a}^{\sqrt{2}}}-{{b}^{\sqrt{3}}}} \\ \end{align}$

$\begin{align} & d)\,\sqrt{{{\left( {{x}^{\pi }}+{{y}^{\pi }} \right)}^{2}}-{{\left( {{4}^{\frac{1}{\pi }}}xy \right)}^{\pi }}} \\ & =\sqrt{{{x}^{2\pi }}+2{{x}^{\pi }}{{y}^{\pi }}+{{y}^{2\pi }}-4{{x}^{\pi }}{{y}^{\pi }}} \\ & =\sqrt{{{x}^{2\pi }}-2{{x}^{\pi }}{{y}^{\pi }}+{{y}^{2\pi }}} \\ & =\sqrt{{{\left( {{x}^{\pi }}-{{y}^{\pi }} \right)}^{2}}} \\ & =\left| {{x}^{\pi }}-{{y}^{\pi }} \right| \\ \end{align}$