求解 y=(4x^2-1)^-4

$$y = ( 4 x ^ { 2 } - 1 ) ^ { - 4 }$$

求解 x 的值 (复数求解)

$$x\in \frac{\sqrt{1-iy^{-\frac{1}{4}}}}{2},-\frac{\sqrt{1-iy^{-\frac{1}{4}}}}{2},-\frac{\sqrt{1-y^{-\frac{1}{4}}}}{2},\frac{\sqrt{1-y^{-\frac{1}{4}}}}{2},-\frac{\sqrt{y^{-\frac{1}{4}}+1}}{2},\frac{\sqrt{y^{-\frac{1}{4}}+1}}{2},-\frac{\sqrt{iy^{-\frac{1}{4}}+1}}{2},\frac{\sqrt{iy^{-\frac{1}{4}}+1}}{2}$$

$$y\neq 0$$

求解 y 的值 (复数求解)

$$y=\frac{1}{\left(4x^{2}-1\right)^{4}}$$

$$x\neq -\frac{1}{2}\text{ and }x\neq \frac{1}{2}$$

求解 x 的值

$$\left\{\begin{matrix}x=-\frac{\sqrt{\sqrt[4]{y}-1}}{2\sqrt[8]{y}}\text{; }x=\frac{\sqrt{\sqrt[4]{y}-1}}{2\sqrt[8]{y}}\text{, }&y\geq 1\\x=-\frac{\sqrt{\sqrt[4]{y}+1}}{2\sqrt[8]{y}}\text{; }x=\frac{\sqrt{\sqrt[4]{y}+1}}{2\sqrt[8]{y}}\text{, }&y>0\end{matrix}\right.$$

求解 y 的值

$$y=\frac{1}{\left(4x^{2}-1\right)^{4}}$$

$$|x|\neq \frac{1}{2}$$