Cho \(a>0,\) biểu thức \(Q=\frac{a^{\frac{4}{3}}\left(a^{-\frac{1}{3}}+a^{\frac{2}{3}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}}+a^{-\frac{1}{4}}\right)}\) được rút gọn bằng
\(\frac{1}{a+1}\). \(\frac{1}{a}\). \(a+1\). \(a\). Hướng dẫn giải:\(Q=\dfrac{a^{\frac{4}{3}}\left(a^{-\frac{1}{3}}+a^{\frac{2}{3}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}}+a^{-\frac{1}{4}}\right)}=\dfrac{a^{\frac{4}{3}}.a^{\frac{2}{3}}\left(a^{-\frac{1}{3}-\frac{2}{3}}+1\right)}{a^{\frac{1}{4}}.a^{-\frac{1}{4}}\left(a^{\frac{3}{4}+\frac{1}{4}}+1\right)}\).
\(=\dfrac{a^{\frac{4}{3}+\frac{2}{3}}\left(a^{-1}+1\right)}{a^{\frac{1}{4}-\frac{1}{4}}\left(a+1\right)}=\dfrac{a^2\left(\frac{1}{a}+1\right)}{a^0\left(a+1\right)}=\dfrac{a\left(1+a\right)}{a+1}=a\)