求解 x 的值
\left\{\begin{matrix}x=\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\text{, }&\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\geq 0\text{ and }\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{72}+\frac{y}{72}\leq 0\text{ and }y^{4}-2y^{3}+y^{2}+648\geq 0\\x=-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\text{, }&-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\geq 0\text{ and }-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{72}+\frac{y}{72}\leq 0\text{ and }y^{4}-2y^{3}+y^{2}+648\geq 0\\x=\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\text{, }&\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\leq 0\text{ and }\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}}{160}+\frac{9y^{2}}{2080}-\frac{9y}{2080}\geq 0\text{ and }y^{4}-2y^{3}+y^{2}+1440\geq 0\\x=\frac{-3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\text{, }&\frac{-3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\leq 0\text{ and }-\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}}{160}+\frac{9y^{2}}{2080}-\frac{9y}{2080}\geq 0\text{ and }y^{4}-2y^{3}+y^{2}+1440\geq 0\end{matrix}\right.
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