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Câu hỏi: Phương trình ${{\log }_{3}}\left( x+2 \right)+\dfrac{1}{2}{{\log }_{3}}{{\left( x-5 \right)}^{2}}+{{\log }_{\dfrac{1}{3}}}8=0$ có bao nhiêu nghiệm thực?
A. 2.
B. 1.
C. 4.
D. 3.
A. 2.
B. 1.
C. 4.
D. 3.
Điều kiện: $\left\{ \begin{aligned}
& x>-2 \\
& x\ne 5 \\
\end{aligned} \right.$.
${{\log }_{3}}\left( x+2 \right)+\dfrac{1}{2}{{\log }_{3}}{{\left( x-5 \right)}^{2}}+{{\log }_{\dfrac{1}{3}}}8=0\Leftrightarrow {{\log }_{3}}\left( x+2 \right)+{{\log }_{3}}\left| x-5 \right|={{\log }_{3}}8$
$\Leftrightarrow {{\log }_{3}}\left[ \left( x+2 \right)\left| x-5 \right| \right]={{\log }_{3}}8\Leftrightarrow \left( x+2 \right)\left| x-5 \right|=8$.
TH1: $x>5$, ta có: $\left( x+2 \right)\left( x-5 \right)=8\Leftrightarrow \left[ \begin{aligned}
& x=6 \left( tm \right) \\
& x=-3 \left( l \right) \\
\end{aligned} \right.$.
TH2: $-2<x<5$, ta có: $\left( x+2 \right)\left( x-5 \right)=-8\Leftrightarrow x=\dfrac{3\pm \sqrt{17}}{2} (tm)$.
& x>-2 \\
& x\ne 5 \\
\end{aligned} \right.$.
${{\log }_{3}}\left( x+2 \right)+\dfrac{1}{2}{{\log }_{3}}{{\left( x-5 \right)}^{2}}+{{\log }_{\dfrac{1}{3}}}8=0\Leftrightarrow {{\log }_{3}}\left( x+2 \right)+{{\log }_{3}}\left| x-5 \right|={{\log }_{3}}8$
$\Leftrightarrow {{\log }_{3}}\left[ \left( x+2 \right)\left| x-5 \right| \right]={{\log }_{3}}8\Leftrightarrow \left( x+2 \right)\left| x-5 \right|=8$.
TH1: $x>5$, ta có: $\left( x+2 \right)\left( x-5 \right)=8\Leftrightarrow \left[ \begin{aligned}
& x=6 \left( tm \right) \\
& x=-3 \left( l \right) \\
\end{aligned} \right.$.
TH2: $-2<x<5$, ta có: $\left( x+2 \right)\left( x-5 \right)=-8\Leftrightarrow x=\dfrac{3\pm \sqrt{17}}{2} (tm)$.
Đáp án D.