(xy-1)^2=(3y+2)(y-2) Denklemini Çözme

$$( x y - 1 ) ^ { 2 } = ( 3 y + 2 ) ( y - 2 )$$

x için çözün (complex solution)

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

$$x=\frac{\sqrt{y-2}\sqrt{3y+2}+1}{y}\text{, }y\neq 0$$

y için çözün (complex solution)

$$\left\{\begin{matrix}y=\frac{\sqrt{19-4x-4x^{2}}+x-2}{x^{2}-3}\text{; }y=\frac{-\sqrt{19-4x-4x^{2}}+x-2}{x^{2}-3}\text{, }&x\neq -\sqrt{3}\text{ and }x\neq \sqrt{3}\\y=-\frac{5}{2\left(2-x\right)}\text{, }&x=-\sqrt{3}\text{ or }x=\sqrt{3}\end{matrix}\right,$$

x için çözün

$$\left\{\begin{matrix}x=-\frac{\sqrt{\left(y-2\right)\left(3y+2\right)}-1}{y}\text{; }x=\frac{\sqrt{\left(y-2\right)\left(3y+2\right)}+1}{y}\text{, }&y\geq 2\\x=\frac{-\sqrt{-\left(2-y\right)\left(3y+2\right)}+1}{y}\text{; }x=\frac{\sqrt{-\left(2-y\right)\left(3y+2\right)}+1}{y}\text{, }&y\leq -\frac{2}{3}\end{matrix}\right,$$

y için çözün

$$\left\{\begin{matrix}y=\frac{\sqrt{19-4x-4x^{2}}+x-2}{x^{2}-3}\text{; }y=\frac{-\sqrt{19-4x-4x^{2}}+x-2}{x^{2}-3}\text{, }&x\geq -\sqrt{5}-\frac{1}{2}\text{ and }x\leq \sqrt{5}-\frac{1}{2}\text{ and }|x|\neq \sqrt{3}\\y=-\frac{5}{2\left(2-x\right)}\text{, }&|x|=\sqrt{3}\end{matrix}\right,$$