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Câu hỏi: Biết phương trình ${{\log }_{3}}x-{{\log }_{5}}x{{\log }_{2}}x=0$ có hai nghiệm phân biệt ${{x}_{1}};{{x}_{2}}$. Tính giá trị biểu thức $T={{\log }_{2}}\left( {{x}_{1}}{{x}_{2}} \right)$.
A. ${{\log }_{5}}2$.
B. ${{\log }_{5}}3$.
C. ${{\log }_{3}}5$.
D. $1+{{\log }_{2}}5$.
A. ${{\log }_{5}}2$.
B. ${{\log }_{5}}3$.
C. ${{\log }_{3}}5$.
D. $1+{{\log }_{2}}5$.
Điều kiện : $x>0$.
Ta có: ${{\log }_{3}}x-{{\log }_{5}}x{{\log }_{2}}x=0$
$\begin{aligned}
& \Leftrightarrow {{\log }_{3}}2{{\log }_{2}}x-{{\log }_{5}}x{{\log }_{2}}x=0 \\
& \Leftrightarrow {{\log }_{2}}x\left( {{\log }_{3}}2-{{\log }_{5}}x \right)=0 \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}x=0 \\
& {{\log }_{3}}2-{{\log }_{5}}x=0 \\
\end{aligned} \right. \\
& \Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x={{5}^{{{\log }_{3}}2}} \\
\end{aligned} \right. \\
\end{aligned}$
Suy ra $T={{\log }_{2}}\left( {{5}^{{{\log }_{3}}2}} \right)={{\log }_{3}}2{{\log }_{2}}5={{\log }_{3}}5$.
Ta có: ${{\log }_{3}}x-{{\log }_{5}}x{{\log }_{2}}x=0$
$\begin{aligned}
& \Leftrightarrow {{\log }_{3}}2{{\log }_{2}}x-{{\log }_{5}}x{{\log }_{2}}x=0 \\
& \Leftrightarrow {{\log }_{2}}x\left( {{\log }_{3}}2-{{\log }_{5}}x \right)=0 \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}x=0 \\
& {{\log }_{3}}2-{{\log }_{5}}x=0 \\
\end{aligned} \right. \\
& \Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x={{5}^{{{\log }_{3}}2}} \\
\end{aligned} \right. \\
\end{aligned}$
Suy ra $T={{\log }_{2}}\left( {{5}^{{{\log }_{3}}2}} \right)={{\log }_{3}}2{{\log }_{2}}5={{\log }_{3}}5$.
Đáp án C.