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Câu hỏi: Tập xác định của hàm số $y=\ln \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2 \right)$ là:
A. $\mathbb{R}\backslash \left\{ -1;0;1 \right\}.$
B. $\left( 0;1 \right).$
C. $\mathbb{R}\backslash \left\{ 0 \right\}.$
D. $\left( 1;+\infty \right).$
A. $\mathbb{R}\backslash \left\{ -1;0;1 \right\}.$
B. $\left( 0;1 \right).$
C. $\mathbb{R}\backslash \left\{ 0 \right\}.$
D. $\left( 1;+\infty \right).$
Điều kiện: $\left\{ \begin{aligned}
& {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-\dfrac{1}{x} \right)}^{2}}>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x\ne \dfrac{1}{x} \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow x\ne \left\{ 1;-1;0 \right\}.$
& {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x-\dfrac{1}{x} \right)}^{2}}>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x\ne \dfrac{1}{x} \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow x\ne \left\{ 1;-1;0 \right\}.$
Đáp án A.