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Câu hỏi: Cho $0<a\ne 1;0<b\ne 1$. Giá trị biểu thức $P={{\log }_{{{a}^{2}}}}\left( {{a}^{10}}{{b}^{2}} \right)+{{\log }_{\sqrt{a}}}\left( \dfrac{a}{\sqrt{b}} \right)$ bằng
A. $7$.
B. $\sqrt{2}$.
C. $\sqrt{3}$.
D. $2$.
A. $7$.
B. $\sqrt{2}$.
C. $\sqrt{3}$.
D. $2$.
Ta có
$\begin{aligned}
& P={{\log }_{{{a}^{2}}}}\left( {{a}^{10}}{{b}^{2}} \right)+{{\log }_{\sqrt{a}}}\left( \dfrac{a}{\sqrt{b}} \right) \\
& =\dfrac{1}{2}{{\log }_{a}}\left( {{a}^{10}}{{b}^{2}} \right)+2{{\log }_{a}}\left( \dfrac{a}{\sqrt{b}} \right) \\
& =\dfrac{1}{2}\left[ 10+2{{\log }_{a}}b \right]+2\left[ 1-\dfrac{1}{2}{{\log }_{a}}b \right] \\
& =5+{{\log }_{a}}b+2-{{\log }_{a}}b \\
& =7 \\
\end{aligned}$
$\begin{aligned}
& P={{\log }_{{{a}^{2}}}}\left( {{a}^{10}}{{b}^{2}} \right)+{{\log }_{\sqrt{a}}}\left( \dfrac{a}{\sqrt{b}} \right) \\
& =\dfrac{1}{2}{{\log }_{a}}\left( {{a}^{10}}{{b}^{2}} \right)+2{{\log }_{a}}\left( \dfrac{a}{\sqrt{b}} \right) \\
& =\dfrac{1}{2}\left[ 10+2{{\log }_{a}}b \right]+2\left[ 1-\dfrac{1}{2}{{\log }_{a}}b \right] \\
& =5+{{\log }_{a}}b+2-{{\log }_{a}}b \\
& =7 \\
\end{aligned}$
Đáp án A.
