求解 {l}{|x^2-1|<1}{xneqpm1}

$$\left\{ \begin{array} { l } { | x ^ { 2 } - 1 | < 1 } \\ { x \neq \pm 1 } \end{array} \right.$$

求值 (复数求解)

$$|x^{2}-1|<1,\ x\neq ±1$$

求解 x 的值

$$\left\{\begin{matrix}x\in \left(-\sqrt{2},0\right)\text{, }&±1\geq 0\text{ or }±1+\sqrt{2}\leq 0\\x\in \left(0,±1\right)\cup \left(±1,\sqrt{2}\right)\text{, }&±1>0\text{ and }±1-\sqrt{2}<0\\x\in \left(0,\sqrt{2}\right)\text{, }&±1-\sqrt{2}\geq 0\text{ or }±1\leq 0\\x\in \left(-\sqrt{2},±1\right)\cup \left(±1,0\right)\text{, }&±1+\sqrt{2}>0\text{ and }±1<0\end{matrix}\right.$$