求解 3x^2+y/2y^2+1=1

$$\frac { 3 x ^ { 2 } + y } { 2 y ^ { 2 } + 1 } = 1$$

求解 x 的值 (复数求解)

$$x=-\frac{\sqrt{6y^{2}-3y+3}}{3}$$

$$x=\frac{\sqrt{6y^{2}-3y+3}}{3}\text{, }y\neq -\frac{\sqrt{2}i}{2}\text{ and }y\neq \frac{\sqrt{2}i}{2}$$

求解 y 的值 (复数求解)

$$\left\{\begin{matrix}y=\frac{\sqrt{24x^{2}-7}+1}{4}\text{, }&x\neq \sqrt{3}\sqrt[4]{2}\left(\frac{1}{6}-\frac{1}{6}i\right)\text{ and }x\neq \sqrt{3}\sqrt[4]{2}\left(-\frac{1}{6}+\frac{1}{6}i\right)\\y=\frac{-\sqrt{24x^{2}-7}+1}{4}\text{, }&x\neq \sqrt{3}\sqrt[4]{2}\left(\frac{1}{6}+\frac{1}{6}i\right)\text{ and }x\neq \sqrt{3}\sqrt[4]{2}\left(-\frac{1}{6}-\frac{1}{6}i\right)\end{matrix}\right.$$

求解 x 的值

$$x=\frac{\sqrt{6y^{2}-3y+3}}{3}$$

$$x=-\frac{\sqrt{6y^{2}-3y+3}}{3}$$

求解 y 的值

$$y=\frac{-\sqrt{24x^{2}-7}+1}{4}$$

$$y=\frac{\sqrt{24x^{2}-7}+1}{4}\text{, }|x|\geq \frac{\sqrt{42}}{12}$$