y=x^2+y^2x/2 шешу

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

x мәнін табыңыз (complex solution)

$$x=\frac{-\sqrt{y^{4}+16y}-y^{2}}{4}$$

$$x=\frac{\sqrt{y^{4}+16y}-y^{2}}{4}$$

y мәнін табыңыз (complex solution)

$$\left\{\begin{matrix}y=-\frac{\sqrt{1-2x^{3}}-1}{x}\text{; }y=\frac{\sqrt{1-2x^{3}}+1}{x}\text{, }&x\neq 0\\y=0\text{, }&x=0\end{matrix}\right.$$

x мәнін табыңыз

$$x=\frac{-\sqrt{y^{4}+16y}-y^{2}}{4}$$

$$x=\frac{\sqrt{y^{4}+16y}-y^{2}}{4}\text{, }y\leq -2^{\frac{4}{3}}\text{ or }y\geq 0$$

y мәнін табыңыз

$$\left\{\begin{matrix}y=-\frac{\sqrt{1-2x^{3}}-1}{x}\text{; }y=\frac{\sqrt{1-2x^{3}}+1}{x}\text{, }&x\neq 0\text{ and }x\leq \frac{2^{\frac{2}{3}}}{2}\\y=0\text{, }&x=0\end{matrix}\right.$$