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Câu hỏi: Cho các số thực $a,b,c$ thuộc khoảng $\left( 1;+\infty \right)$ và $\log _{\sqrt{a}}^{2}b+{{\log }_{b}}c.{{\log }_{b}}\dfrac{{{c}^{2}}}{b}+9{{\log }_{a}}c=4{{\log }_{a}}b$.
Giá trị của biểu thức ${{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}$ bằng
A. $\dfrac{1}{2}$.
B. $1$.
C. $2$.
D. $3$.
Giá trị của biểu thức ${{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}$ bằng
A. $\dfrac{1}{2}$.
B. $1$.
C. $2$.
D. $3$.
$\begin{aligned}
& \log _{\sqrt{a}}^{2}b+{{\log }_{b}}c.{{\log }_{b}}\dfrac{{{c}^{2}}}{b}+9{{\log }_{a}}c=4{{\log }_{a}}b \\
& \Leftrightarrow 4\log _{a}^{2}b+\dfrac{1}{2}{{\log }_{b}}{{c}^{2}}\left( {{\log }_{b}}{{c}^{2}}-1 \right)+\dfrac{9}{2}{{\log }_{a}}{{c}^{2}}-4{{\log }_{a}}b=0 \\
& \Leftrightarrow 8\log _{a}^{2}b+\log _{b}^{2}{{c}^{2}}-{{\log }_{b}}{{c}^{2}}+9{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}-8{{\log }_{a}}b=0 \\
& \Leftrightarrow 8\log _{a}^{2}b-8{{\log }_{a}}b+8{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}+\log _{b}^{2}{{c}^{2}}-{{\log }_{b}}{{c}^{2}}+{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}=0 \\
& \Leftrightarrow 8{{\log }_{a}}b\left( {{\log }_{a}}b-1+{{\log }_{b}}{{c}^{2}} \right)+{{\log }_{b}}{{c}^{2}}\left( {{\log }_{b}}{{c}^{2}}-1+{{\log }_{a}}b \right)=0 \\
& \Leftrightarrow \left( {{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}-1 \right)\left( 8{{\log }_{a}}b+{{\log }_{b}}{{c}^{2}} \right)=0 \\
& \Leftrightarrow \left[ \begin{aligned}
& {{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=1 \\
& 8{{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=0 \\
\end{aligned} \right. \\
\end{aligned}$
Vậy ${{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=1$.
& \log _{\sqrt{a}}^{2}b+{{\log }_{b}}c.{{\log }_{b}}\dfrac{{{c}^{2}}}{b}+9{{\log }_{a}}c=4{{\log }_{a}}b \\
& \Leftrightarrow 4\log _{a}^{2}b+\dfrac{1}{2}{{\log }_{b}}{{c}^{2}}\left( {{\log }_{b}}{{c}^{2}}-1 \right)+\dfrac{9}{2}{{\log }_{a}}{{c}^{2}}-4{{\log }_{a}}b=0 \\
& \Leftrightarrow 8\log _{a}^{2}b+\log _{b}^{2}{{c}^{2}}-{{\log }_{b}}{{c}^{2}}+9{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}-8{{\log }_{a}}b=0 \\
& \Leftrightarrow 8\log _{a}^{2}b-8{{\log }_{a}}b+8{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}+\log _{b}^{2}{{c}^{2}}-{{\log }_{b}}{{c}^{2}}+{{\log }_{a}}b.{{\log }_{b}}{{c}^{2}}=0 \\
& \Leftrightarrow 8{{\log }_{a}}b\left( {{\log }_{a}}b-1+{{\log }_{b}}{{c}^{2}} \right)+{{\log }_{b}}{{c}^{2}}\left( {{\log }_{b}}{{c}^{2}}-1+{{\log }_{a}}b \right)=0 \\
& \Leftrightarrow \left( {{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}-1 \right)\left( 8{{\log }_{a}}b+{{\log }_{b}}{{c}^{2}} \right)=0 \\
& \Leftrightarrow \left[ \begin{aligned}
& {{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=1 \\
& 8{{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=0 \\
\end{aligned} \right. \\
\end{aligned}$
Vậy ${{\log }_{a}}b+{{\log }_{b}}{{c}^{2}}=1$.
Đáp án B.