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Câu hỏi: Cho số thực $a$ thỏa mãn $0<a\ne 1.$ Tính giá trị của biểu thức $T={{\log }_{a}}\dfrac{{{a}^{2}}.\sqrt[3]{a}.\sqrt[5]{{{a}^{3}}}}{\sqrt[15]{{{a}^{4}}}}.$
A. $T=8$
B. $T=11$
C. $T=\dfrac{8}{3}$
D. $T=\dfrac{17}{15}$
A. $T=8$
B. $T=11$
C. $T=\dfrac{8}{3}$
D. $T=\dfrac{17}{15}$
Cách giải:
$T={{\log }_{a}}\dfrac{{{a}^{2}}\sqrt[3]{a}.\sqrt[5]{{{a}^{3}}}}{\sqrt[15]{{{a}^{4}}}}={{\log }_{a}}\dfrac{{{a}^{2}}.{{a}^{\dfrac{1}{3}}}.{{a}^{\dfrac{3}{5}}}}{{{a}^{\dfrac{4}{15}}}}$
$={{\log }_{a}}\dfrac{{{a}^{2+\dfrac{1}{3}+\dfrac{3}{5}}}}{{{a}^{\dfrac{4}{15}}}}={{\log }_{a}}\dfrac{{{a}^{\dfrac{44}{15}}}}{{{a}^{\dfrac{4}{15}}}}={{\log }_{a}}{{a}^{\dfrac{44}{15}-\dfrac{4}{15}}}$
$={{\log }_{a}}{{a}^{\dfrac{8}{3}}}=\dfrac{8}{3}$
$T={{\log }_{a}}\dfrac{{{a}^{2}}\sqrt[3]{a}.\sqrt[5]{{{a}^{3}}}}{\sqrt[15]{{{a}^{4}}}}={{\log }_{a}}\dfrac{{{a}^{2}}.{{a}^{\dfrac{1}{3}}}.{{a}^{\dfrac{3}{5}}}}{{{a}^{\dfrac{4}{15}}}}$
$={{\log }_{a}}\dfrac{{{a}^{2+\dfrac{1}{3}+\dfrac{3}{5}}}}{{{a}^{\dfrac{4}{15}}}}={{\log }_{a}}\dfrac{{{a}^{\dfrac{44}{15}}}}{{{a}^{\dfrac{4}{15}}}}={{\log }_{a}}{{a}^{\dfrac{44}{15}-\dfrac{4}{15}}}$
$={{\log }_{a}}{{a}^{\dfrac{8}{3}}}=\dfrac{8}{3}$
Đáp án C.