Giải bài 2.3 trang 100 - SBT Giải tích lớp 12
Cho a và b là các số dương. Đơn giản các biểu thức sau:
\(\begin{align} & a)\,\frac{{{a}^{\frac{4}{3}}}\left( {{a}^{-\frac{1}{3}}}+{{a}^{\frac{2}{3}}} \right)}{{{a}^{\frac{1}{4}}}\left( {{a}^{\frac{3}{4}}}+{{a}^{-\frac{1}{4}}} \right)}; \\ & b)\,\frac{{{a}^{\frac{1}{3}}}\sqrt{b}+{{b}^{\frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}; \\ & c)\,\left( \sqrt[3]{a}+\sqrt[3]{b} \right)\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}}-\sqrt[3]{ab} \right); \\ & d)\,\left( {{a}^{\frac{1}{3}}}+{{b}^{\frac{1}{3}}} \right):\left( 2+\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \right) \\ \end{align} \)
\(\begin{align}
&a)\,\frac{{{a}^{\frac{4}{3}}}\left( {{a}^{-\frac{1}{3}}}+{{a}^{\frac{2}{3}}} \right)}{{{a}^{\frac{1}{4}}}\left( {{a}^{\frac{3}{4}}}+{{a}^{-\frac{1}{4}}} \right)} \\
& =\frac{{{a}^{\frac{4}{3}}}.{{a}^{\frac{-1}{3}}}+{{a}^{\frac{4}{3}}}.{{a}^{\frac{2}{3}}}}{{{a}^{\frac{1}{4}}}.{{a}^{\frac{3}{4}}}+{{a}^{\frac{1}{4}}}.{{a}^{\frac{-1}{4}}}} \\
& =\frac{{{a}^{\frac{4}{3}+\frac{-1}{3}}}+{{a}^{\frac{4}{3}+\frac{2}{3}}}}{{{a}^{\frac{1}{4}+\frac{3}{4}}}+{{a}^{\frac{1}{4}+\frac{-1}{4}}}} \\
& =\frac{a+{{a}^{2}}}{a+1} \\
& =\frac{a\left( 1+a \right)}{1+a}=a \\
\end{align}\)
\(b)\,\dfrac{{{a}^{\frac{1}{3}}}\sqrt{b}+{{b}^{\frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\\
=\dfrac{{{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{2}}}+{{b}^{\frac{1}{3}}}.{{a}^{\frac{1}{2}}}}{{{a}^{\frac{1}{6}}}+{{b}^{\frac{1}{6}}}}\\
=\dfrac{{{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{3}}}\left( {{b}^{\frac{1}{6}}}+{{a}^{\frac{1}{6}}} \right)}{{{a}^{\frac{1}{6}}}+{{b}^{\frac{1}{6}}}}\\
={{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{3}}}=\sqrt[3]{ab} \)
\(c)\,\left( \sqrt[3]{a}+\sqrt[3]{b} \right)\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}}-\sqrt[3]{ab} \right)\\
\begin{align}
& =\left( {{a}^{\frac{1}{3}}}+{{b}^{\frac{1}{3}}} \right)\left[ {{\left( {{a}^{\frac{1}{3}}} \right)}^{2}}-{{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{3}}}-{{\left( {{b}^{\frac{1}{3}}} \right)}^{2}} \right] \\
& ={{\left( {{a}^{\frac{1}{3}}} \right)}^{3}}+{{\left( {{b}^{\frac{1}{3}}} \right)}^{3}}=a+b \\
\end{align} \)
\(d)\,\left( {{a}^{\frac{1}{3}}}+{{b}^{\frac{1}{3}}} \right):\left( 2+\sqrt[3]{\dfrac{a}{b}}+\sqrt[3]{\dfrac{b}{a}} \right)\\
\begin{align}
& =\dfrac{{{a}^{\frac{1}{3}}}+{{b}^{\frac{1}{3}}}}{\dfrac{2{{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{3}}}+{{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}}}{{{a}^{\frac{1}{3}}}.{{b}^{\frac{1}{3}}}}} \\
& =\frac{\sqrt[3]{ab}\left( \sqrt[3]{a}+\sqrt[3]{b} \right)}{{{\left( \sqrt[3]{a}+\sqrt[3]{b} \right)}^{2}}} \\
& =\frac{\sqrt[3]{ab}}{\sqrt[3]{a}+\sqrt[3]{b}} \\
\end{align}\)