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Câu hỏi: Phương trình $\ln \left( x-\dfrac{2}{3} \right)\ln \left( x+\dfrac{2}{3} \right)\ln \left( x+\dfrac{1}{3} \right)\ln \left( x+\dfrac{1}{6} \right)=0$ có bao nhiêu nghiệm thực.
A. $3$.
B. $4$.
C. $2$.
D. $1$.
A. $3$.
B. $4$.
C. $2$.
D. $1$.
Đk: $x>\dfrac{2}{3}.$
Khi đó, $\ln \left( x-\dfrac{2}{3} \right)\ln \left( x+\dfrac{2}{3} \right)\ln \left( x+\dfrac{1}{3} \right)\ln \left( x+\dfrac{1}{6} \right)=0$
$\Leftrightarrow \left[ \begin{aligned}
& \ln \left( x-\dfrac{2}{3} \right)=0\Leftrightarrow x=\dfrac{5}{3} \left( thoa man \right) \\
& \ln \left( x+\dfrac{2}{3} \right)=0\Leftrightarrow x=\dfrac{1}{3} \left( loai \right) \\
& \ln \left( x+\dfrac{1}{3} \right)=0\Leftrightarrow x=\dfrac{2}{3}\left( loai \right) \\
& \ln \left( x+\dfrac{1}{6} \right)=0\Leftrightarrow x=\dfrac{5}{6}\left( thoa man \right) \\
\end{aligned} \right.$
Vậy phương trình đã cho có $2$ nghiệm thực.
Khi đó, $\ln \left( x-\dfrac{2}{3} \right)\ln \left( x+\dfrac{2}{3} \right)\ln \left( x+\dfrac{1}{3} \right)\ln \left( x+\dfrac{1}{6} \right)=0$
$\Leftrightarrow \left[ \begin{aligned}
& \ln \left( x-\dfrac{2}{3} \right)=0\Leftrightarrow x=\dfrac{5}{3} \left( thoa man \right) \\
& \ln \left( x+\dfrac{2}{3} \right)=0\Leftrightarrow x=\dfrac{1}{3} \left( loai \right) \\
& \ln \left( x+\dfrac{1}{3} \right)=0\Leftrightarrow x=\dfrac{2}{3}\left( loai \right) \\
& \ln \left( x+\dfrac{1}{6} \right)=0\Leftrightarrow x=\dfrac{5}{6}\left( thoa man \right) \\
\end{aligned} \right.$
Vậy phương trình đã cho có $2$ nghiệm thực.
Đáp án C.