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Câu hỏi: Tìm tập xác định $D$ của hàm số $y=\dfrac{1}{{{\log }_{3}}\left( 2{{x}^{2}}-x \right)}$ .
A. $D=\left( -\infty \ ;\ 0 \right]\cup \left[ \dfrac{1}{2}\ ;\ +\infty \right)$.
B. $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$.
C. $D=\left( -\infty \ ;\ 0 \right]\cup \left[ \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$.
D. $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)$.
A. $D=\left( -\infty \ ;\ 0 \right]\cup \left[ \dfrac{1}{2}\ ;\ +\infty \right)$.
B. $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$.
C. $D=\left( -\infty \ ;\ 0 \right]\cup \left[ \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$.
D. $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)$.
Hàm số xác định khi $\left\{ \begin{aligned}
& 2{{x}^{2}}-x>0 \\
& \log {}_{3}\left( 2{{x}^{2}}-x \right)\ne 0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x<0\vee x>\dfrac{1}{2} \\
& 2{{x}^{2}}-x\ne 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<0\vee x>\dfrac{1}{2} \\
& x\ne -\dfrac{1}{2} \\
& x\ne 1 \\
\end{aligned} \right.$.
Vậy $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$ .
& 2{{x}^{2}}-x>0 \\
& \log {}_{3}\left( 2{{x}^{2}}-x \right)\ne 0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x<0\vee x>\dfrac{1}{2} \\
& 2{{x}^{2}}-x\ne 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<0\vee x>\dfrac{1}{2} \\
& x\ne -\dfrac{1}{2} \\
& x\ne 1 \\
\end{aligned} \right.$.
Vậy $D=\left( -\infty \ ;\ 0 \right)\cup \left( \dfrac{1}{2}\ ;\ +\infty \right)\backslash \left\{ -\dfrac{1}{2}\ ;\ 1 \right\}$ .
Đáp án B.