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Câu hỏi: Cho các số dương $a,b,c$ khác 1 thỏa mãn ${{\log }_{a}}\left( bc \right)=3,{{\log }_{b}}\left( ca \right)=4.$ Tính giá trị của ${{\log }_{c}}\left( ab \right).$
A. $\dfrac{16}{9}.$
B. $\dfrac{16}{4}.$
C. $\dfrac{11}{9}.$
D. $\dfrac{9}{11}.$
A. $\dfrac{16}{9}.$
B. $\dfrac{16}{4}.$
C. $\dfrac{11}{9}.$
D. $\dfrac{9}{11}.$
Ta có:
${{\log }_{a}}\left( bc \right)=\dfrac{{{\log }_{c}}\left( bc \right)}{{{\log }_{c}}a}=\dfrac{{{\log }_{c}}b+1}{{{\log }_{c}}a}=3\Rightarrow 3{{\log }_{c}}a-{{\log }_{c}}b=1.\left( 1 \right)$
${{\log }_{b}}\left( ca \right)=\dfrac{{{\log }_{c}}\left( ca \right)}{{{\log }_{c}}b}=\dfrac{{{\log }_{c}}a+1}{{{\log }_{c}}b}=4\Rightarrow {{\log }_{c}}a-4{{\log }_{c}}b=-1.\left( 2 \right)$
Từ (1) và (2) ta có hệ phương trình
$\left\{ \begin{aligned}
& 3{{\log }_{c}}a-{{\log }_{c}}b=1 \\
& {{\log }_{c}}a-4{{\log }_{c}}b=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{c}}a=\dfrac{5}{11} \\
& {{\log }_{c}}b=\dfrac{4}{11} \\
\end{aligned} \right.\Rightarrow {{\log }_{c}}\left( ab \right)={{\log }_{c}}a+{{\log }_{c}}b=\dfrac{9}{11}.$
${{\log }_{a}}\left( bc \right)=\dfrac{{{\log }_{c}}\left( bc \right)}{{{\log }_{c}}a}=\dfrac{{{\log }_{c}}b+1}{{{\log }_{c}}a}=3\Rightarrow 3{{\log }_{c}}a-{{\log }_{c}}b=1.\left( 1 \right)$
${{\log }_{b}}\left( ca \right)=\dfrac{{{\log }_{c}}\left( ca \right)}{{{\log }_{c}}b}=\dfrac{{{\log }_{c}}a+1}{{{\log }_{c}}b}=4\Rightarrow {{\log }_{c}}a-4{{\log }_{c}}b=-1.\left( 2 \right)$
Từ (1) và (2) ta có hệ phương trình
$\left\{ \begin{aligned}
& 3{{\log }_{c}}a-{{\log }_{c}}b=1 \\
& {{\log }_{c}}a-4{{\log }_{c}}b=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{c}}a=\dfrac{5}{11} \\
& {{\log }_{c}}b=\dfrac{4}{11} \\
\end{aligned} \right.\Rightarrow {{\log }_{c}}\left( ab \right)={{\log }_{c}}a+{{\log }_{c}}b=\dfrac{9}{11}.$
Đáp án D.