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Câu hỏi: Cho hàm số $f\left( a \right)=\dfrac{{{a}^{\dfrac{2}{3}}}\left( \sqrt[3]{{{a}^{-1}}}-\sqrt[3]{a} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}$ với $a>0,a\ne 1a$, Tính giá trị $f\left( {{2019}^{2018}} \right)$.
A. ${{2019}^{1009}}$.
B. ${{2019}^{1009}}+1$.
C. $-{{2019}^{1009}}+1$.
D. $-{{2019}^{1009}}-1$.
A. ${{2019}^{1009}}$.
B. ${{2019}^{1009}}+1$.
C. $-{{2019}^{1009}}+1$.
D. $-{{2019}^{1009}}-1$.
Ta có
$f\left( a \right)=\dfrac{{{a}^{\dfrac{2}{3}}}\left( \sqrt[3]{{{a}^{-2}}}-\sqrt[3]{a} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}=\dfrac{{{a}^{\dfrac{2}{3}}}\left( {{a}^{\dfrac{-2}{3}}}-{{a}^{\dfrac{1}{3}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( {{a}^{\dfrac{3}{8}}}-{{a}^{\dfrac{-1}{8}}} \right)}=\dfrac{1-a}{{{a}^{\dfrac{1}{2}}}-1}=\dfrac{-\left( {{a}^{\dfrac{1}{2}}}-1 \right)\left( {{a}^{\dfrac{1}{2}}}+1 \right)}{{{a}^{\dfrac{1}{2}}}-1}=-{{a}^{\dfrac{1}{2}}}-1$.
Khi đó $f\left( {{2019}^{2018}} \right)=-{{\left( {{2019}^{2018}} \right)}^{\dfrac{1}{2}}}-1=-{{2019}^{1009}}-1$.
$f\left( a \right)=\dfrac{{{a}^{\dfrac{2}{3}}}\left( \sqrt[3]{{{a}^{-2}}}-\sqrt[3]{a} \right)}{{{a}^{\dfrac{1}{8}}}\left( \sqrt[8]{{{a}^{3}}}-\sqrt[8]{{{a}^{-1}}} \right)}=\dfrac{{{a}^{\dfrac{2}{3}}}\left( {{a}^{\dfrac{-2}{3}}}-{{a}^{\dfrac{1}{3}}} \right)}{{{a}^{\dfrac{1}{8}}}\left( {{a}^{\dfrac{3}{8}}}-{{a}^{\dfrac{-1}{8}}} \right)}=\dfrac{1-a}{{{a}^{\dfrac{1}{2}}}-1}=\dfrac{-\left( {{a}^{\dfrac{1}{2}}}-1 \right)\left( {{a}^{\dfrac{1}{2}}}+1 \right)}{{{a}^{\dfrac{1}{2}}}-1}=-{{a}^{\dfrac{1}{2}}}-1$.
Khi đó $f\left( {{2019}^{2018}} \right)=-{{\left( {{2019}^{2018}} \right)}^{\dfrac{1}{2}}}-1=-{{2019}^{1009}}-1$.
Đáp án D.