Cho hàm số f(a)=a^-1/3*(căn bậc 3(a)-căn bậc 3(a^4))/a^1/8*( căn bậc 8(a^3-căn bậc 8(a^-1)) với

* Hướng dẫn giải

Ta có: \(f\left( a \right) = \frac{{{a^{\frac{{ - 1}}{3}}}\left( {\sqrt[3]{a} - \sqrt[3]{{{a^4}}}} \right)}}{{{a^{\frac{1}{8}}}\left( {\sqrt[8]{{{a^3}}} - \sqrt[8]{{{a^{ - 1}}}}} \right)}} = \frac{{{a^{\frac{{ - 1}}{3}}}\left( {{a^{\frac{1}{3}}} - {a^{\frac{4}{3}}}} \right)}}{{{a^{\frac{1}{8}}}\left( {{a^{\frac{3}{8}}} - {a^{\frac{{ - 1}}{8}}}} \right)}} = \frac{{{a^{\frac{{ - 1}}{3}}}{a^{\frac{1}{3}}}\left( {1 - a} \right)}}{{{a^{\frac{1}{8}}}{a^{\frac{{ - 1}}{8}}}\left( {{a^{\frac{1}{2}}} - 1} \right)}} = \frac{{\left( {1 - a} \right)}}{{\sqrt a - 1}} = - \sqrt a - 1\)

$\Rightarrow f\left({2021}^{2020}\right)=-{\left({2021}^{2020}\right)}^{\frac{1}{2}}-1=-{2021}^{1010}-1.$

Đáp án D