Rezolvați 9x^3y-6x^2y^2+3xy^3/3xy=3x

$$\frac { 9 x ^ { 3 } y - 6 x ^ { 2 } y ^ { 2 } + 3 x y ^ { 3 } } { 3 x y } = 3 x$$

Rezolvați pentru x (complex solution)

$$x=\frac{\sqrt{9+12y-8y^{2}}}{6}+\frac{y}{3}+\frac{1}{2}$$

$$x=-\frac{\sqrt{9+12y-8y^{2}}}{6}+\frac{y}{3}+\frac{1}{2}\text{, }y\neq 0$$

Rezolvați pentru y (complex solution)

$$\left\{\begin{matrix}y=-\sqrt{x\left(3-2x\right)}+x\text{, }&x\neq 1\text{ and }x\neq 0\\y=\sqrt{x\left(3-2x\right)}+x\text{, }&x\neq 0\end{matrix}\right,$$

Rezolvați pentru x

$$x=\frac{\sqrt{9+12y-8y^{2}}}{6}+\frac{y}{3}+\frac{1}{2}$$

$$x=-\frac{\sqrt{9+12y-8y^{2}}}{6}+\frac{y}{3}+\frac{1}{2}\text{, }y\neq 0\text{ and }y\geq \frac{3-3\sqrt{3}}{4}\text{ and }y\leq \frac{3\sqrt{3}+3}{4}$$

Rezolvați pentru y

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