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Câu hỏi: Hàm số $y={{\left( 4{{x}^{2}}-1 \right)}^{-4}}$ có tập xác định là
A. $\mathbb{R}\backslash \left\{ -\dfrac{1}{2};\dfrac{1}{2} \right\}$.
B. $\mathbb{R}$.
C. $\left( -\infty ;-\dfrac{1}{2} \right)\cup \left( \dfrac{1}{2};+\infty \right)$.
D. $\left( -\dfrac{1}{2};\dfrac{1}{2} \right)$.
Ta có: $y=\dfrac{1}{{{\left( 4{{x}^{2}}-1 \right)}^{4}}}$ có nghĩa khi $4{{x}^{2}}-1\ne 0\Leftrightarrow \left\{ \begin{aligned}
& x\ne \dfrac{1}{2} \\
& x\ne -\dfrac{1}{2} \\
\end{aligned} \right.$.
A. $\mathbb{R}\backslash \left\{ -\dfrac{1}{2};\dfrac{1}{2} \right\}$.
B. $\mathbb{R}$.
C. $\left( -\infty ;-\dfrac{1}{2} \right)\cup \left( \dfrac{1}{2};+\infty \right)$.
D. $\left( -\dfrac{1}{2};\dfrac{1}{2} \right)$.
Ta có: $y=\dfrac{1}{{{\left( 4{{x}^{2}}-1 \right)}^{4}}}$ có nghĩa khi $4{{x}^{2}}-1\ne 0\Leftrightarrow \left\{ \begin{aligned}
& x\ne \dfrac{1}{2} \\
& x\ne -\dfrac{1}{2} \\
\end{aligned} \right.$.
Đáp án A.