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Câu hỏi: Cho hàm số $f\left( a \right)=\dfrac{{{a}^{-\dfrac{1}{3}}}\left( \sqrt[3]{a}-\sqrt[3]{{{a}^{4}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}$ với $a>0,a\ne 1.$ Tính giá trị $M=f\left( {{2021}^{2020}} \right).$
A. $M=1-{{2021}^{2020}}.$
B. $M=-{{2021}^{1010}}-1.$
C. $M={{2021}^{1010}}-1.$
D. $M={{2021}^{2019}}-1.$
A. $M=1-{{2021}^{2020}}.$
B. $M=-{{2021}^{1010}}-1.$
C. $M={{2021}^{1010}}-1.$
D. $M={{2021}^{2019}}-1.$
Phương pháp:
Sử dụng công thức $\sqrt[m]{{{a}^{n}}}={{a}^{\dfrac{n}{m}}},{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}.$
Cách giải:$\begin{aligned}
& f\left( a \right)=\dfrac{{{a}^{-\dfrac{1}{3}}}\left( \sqrt[3]{a}-\sqrt[3]{{{a}^{4}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}=\dfrac{{{a}^{\dfrac{1}{3}}}\left( {{a}^{\dfrac{1}{3}}}-{{a}^{\dfrac{4}{3}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( {{a}^{\dfrac{3}{8}}}-{{a}^{\dfrac{1}{8}}} \right)} \\
& =\dfrac{1-a}{{{a}^{\dfrac{1}{2}}}-1}=\dfrac{1-a}{\sqrt{a}-1}=-\dfrac{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}{\sqrt{a}-1}=-\sqrt{a}-1 \\
& \Rightarrow f\left( {{2021}^{2020}} \right)=-\sqrt{{{2021}^{2020}}}-1=-{{2021}^{1010}}-1. \\
\end{aligned}$
Sử dụng công thức $\sqrt[m]{{{a}^{n}}}={{a}^{\dfrac{n}{m}}},{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}.$
Cách giải:$\begin{aligned}
& f\left( a \right)=\dfrac{{{a}^{-\dfrac{1}{3}}}\left( \sqrt[3]{a}-\sqrt[3]{{{a}^{4}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}=\dfrac{{{a}^{\dfrac{1}{3}}}\left( {{a}^{\dfrac{1}{3}}}-{{a}^{\dfrac{4}{3}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( {{a}^{\dfrac{3}{8}}}-{{a}^{\dfrac{1}{8}}} \right)} \\
& =\dfrac{1-a}{{{a}^{\dfrac{1}{2}}}-1}=\dfrac{1-a}{\sqrt{a}-1}=-\dfrac{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}{\sqrt{a}-1}=-\sqrt{a}-1 \\
& \Rightarrow f\left( {{2021}^{2020}} \right)=-\sqrt{{{2021}^{2020}}}-1=-{{2021}^{1010}}-1. \\
\end{aligned}$
Đáp án B.