Cho hàm số f(a)=a^2/3(căn bậc 3 a^-1-căn bậc ba

* Hướng dẫn giải

Đáp án D

Ta có

$$\begin{gathered} f\left(a\right)=\frac{a^{\frac{2}{3}}\left(\sqrt[3]{a^{-2}}-\sqrt[3]{a}\right)}{a^{\frac{1}{8}}\left(\sqrt[8]{a^3}-\sqrt[8]{a^{-1}}\right)}=\frac{a^{\frac{2}{3}}\left(a^{\frac{-2}{3}}-a^{\frac{1}{3}}\right)}{a^{\frac{1}{8}}\left(a^{\frac{3}{8}}-a^{\frac{-1}{8}}\right)}=\frac{1-a}{a^{\frac{1}{2}}-1}=\frac{-\left(a^{\frac{1}{2}}-1\right)\left(a^{\frac{1}{2}}+1\right)}{a^{\frac{1}{2}}-1} \\ =-a^{\frac{1}{2}}-1 \end{gathered}$$.

Khi đó $f\left({2019}^{2018}\right)=-{\left({2019}^{2018}\right)}^{\frac{1}{2}}-1=-{2019}^{1009}-1$.