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Câu hỏi: Cho $a, b$ là các số thực dương lớn hơn $1$ thỏa mãn ${{\log }_{a}}b=3$. Tính gái trị biểu thức $P={{\log }_{{{a}^{2}}b}}{{a}^{3}}-3{{\log }_{{{a}^{2}}}}2.{{\log }_{4}}\left( \dfrac{a}{b} \right)$.
A. $P=\dfrac{15}{8}$.
B. $P=\dfrac{18}{25}$.
C. $P=\dfrac{21}{10}$.
D. $P=\dfrac{7}{5}$.
A. $P=\dfrac{15}{8}$.
B. $P=\dfrac{18}{25}$.
C. $P=\dfrac{21}{10}$.
D. $P=\dfrac{7}{5}$.
Ta có: ${{\log }_{a}}b=3\Leftrightarrow b={{a}^{3}}$
$P={{\log }_{{{a}^{2}}b}}{{a}^{3}}-3{{\log }_{{{a}^{2}}}}2.{{\log }_{4}}\left( \dfrac{a}{b} \right)={{\log }_{{{a}^{5}}}}{{a}^{3}}-3{{\log }_{{{a}^{2}}}}2.{{\log }_{4}}\left( \dfrac{a}{{{a}^{3}}} \right)$ $=\dfrac{3}{5}-3.\dfrac{1}{2}.{{\log }_{a}}2.{{\log }_{{{2}^{2}}}}\left( \dfrac{1}{{{a}^{2}}} \right)$.
$=\dfrac{3}{5}-3.\dfrac{1}{2}.{{\log }_{a}}2.\dfrac{1}{2}.{{\log }_{2}}{{a}^{-2}}=\dfrac{3}{5}+\dfrac{3}{2}.{{\log }_{a}}2.{{\log }_{2}}a=\dfrac{3}{5}+\dfrac{3}{2}=\dfrac{21}{10}$.
$P={{\log }_{{{a}^{2}}b}}{{a}^{3}}-3{{\log }_{{{a}^{2}}}}2.{{\log }_{4}}\left( \dfrac{a}{b} \right)={{\log }_{{{a}^{5}}}}{{a}^{3}}-3{{\log }_{{{a}^{2}}}}2.{{\log }_{4}}\left( \dfrac{a}{{{a}^{3}}} \right)$ $=\dfrac{3}{5}-3.\dfrac{1}{2}.{{\log }_{a}}2.{{\log }_{{{2}^{2}}}}\left( \dfrac{1}{{{a}^{2}}} \right)$.
$=\dfrac{3}{5}-3.\dfrac{1}{2}.{{\log }_{a}}2.\dfrac{1}{2}.{{\log }_{2}}{{a}^{-2}}=\dfrac{3}{5}+\dfrac{3}{2}.{{\log }_{a}}2.{{\log }_{2}}a=\dfrac{3}{5}+\dfrac{3}{2}=\dfrac{21}{10}$.
Đáp án C.