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Câu hỏi: Tập nghiệm của phương trình $\log (-x+3)-1=\log \left(\dfrac{1}{2}-x\right)$ là
A. $\left\{\dfrac{1}{3} ; \dfrac{2}{9}\right\}$.
B. $\left\{\dfrac{2}{9}\right\}$.
C. $\left\{ -\dfrac{2}{9} \right\}$.
D. $\left\{\dfrac{1}{4}\right\}$
A. $\left\{\dfrac{1}{3} ; \dfrac{2}{9}\right\}$.
B. $\left\{\dfrac{2}{9}\right\}$.
C. $\left\{ -\dfrac{2}{9} \right\}$.
D. $\left\{\dfrac{1}{4}\right\}$
$\log (-x+3)-1=\log \left( \dfrac{1}{2}-x \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \log (-x+3)-\log 10=\log \left( \dfrac{1}{2}-x \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \log \left( \dfrac{-x+3}{10} \right)=\log \left( \dfrac{1}{2}-x \right) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \dfrac{-x+3}{10}=\dfrac{1}{2}-x \\
\end{aligned} \right.\Leftrightarrow x=\dfrac{2}{9}$.
$\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \log (-x+3)-\log 10=\log \left( \dfrac{1}{2}-x \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \log \left( \dfrac{-x+3}{10} \right)=\log \left( \dfrac{1}{2}-x \right) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x<\dfrac{1}{2} \\
& \dfrac{-x+3}{10}=\dfrac{1}{2}-x \\
\end{aligned} \right.\Leftrightarrow x=\dfrac{2}{9}$.
Đáp án B.