Tìm tập xác định $D$ của hàm số $y = \dfrac{1}{\left(log\right)_{3} \left( 2 x^{2} - x \right)}$.

A.  

$D = \left( - \infty \textrm{ } ; \textrm{ } 0 \left]\right. \cup \left[\right. \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right)$.

B.  

$D = \left( - \infty \textrm{ } ; \textrm{ } 0 \right) \cup \left( \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right) \left{ - \dfrac{1}{2} \textrm{ } ; \textrm{ } 1 \right}$.

C.  

$D = \left( - \infty \textrm{ } ; \textrm{ } 0 \left]\right. \cup \left[\right. \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right) \left{ - \dfrac{1}{2} \textrm{ } ; \textrm{ } 1 \right}$.

D.  

$D = \left( - \infty \textrm{ } ; \textrm{ } 0 \right) \cup \left( \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right)$.

Đáp án đúng là: B

Tìm tập xác định $D$ của hàm số $y = \dfrac{1}{\left(log\right)_{3} \left( 2 x^{2} - x \right)}$.
A. $D = \left( - \infty \textrm{ } ; \textrm{ } 0 \left]\right. \cup \left[\right. \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right)$. B. $D = \left( - \infty \textrm{ } ; \textrm{ } 0 \right) \cup \left( \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right) \left{ - \dfrac{1}{2} \textrm{ } ; \textrm{ } 1 \right}$.
C. $D = \left( - \infty \textrm{ } ; \textrm{ } 0 \left]\right. \cup \left[\right. \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right) \left{ - \dfrac{1}{2} \textrm{ } ; \textrm{ } 1 \right}$. D. $D = \left( - \infty \textrm{ } ; \textrm{ } 0 \right) \cup \left( \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right)$.
Lời giải
Hàm số xác định khi $\left{\right. 2 x^{2} - x > 0 \\ log \_{3}^{}\left(\left( 2 x^{2} - x \right)\right) \neq 0$ $\Leftrightarrow \left{\right. x < 0 \lor x > \dfrac{1}{2} \\ 2 x^{2} - x \neq 1 \Leftrightarrow \left{\right. x < 0 \lor x > \dfrac{1}{2} \\ x \neq - \dfrac{1}{2} \\ x \neq 1$.
Vậy $D = \left( - \infty \textrm{ } ; \textrm{ } 0 \right) \cup \left( \dfrac{1}{2} \textrm{ } ; \textrm{ } + \infty \right) \left{ - \dfrac{1}{2} \textrm{ } ; \textrm{ } 1 \right}$.