Решете x(a-b)^2=y

$$x ( a - b ) ^ { 2 } = y$$

Реши за a (complex solution)

$$\left\{\begin{matrix}a=b+x^{-\frac{1}{2}}\sqrt{y}\text{; }a=b-x^{-\frac{1}{2}}\sqrt{y}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&y=0\text{ and }x=0\end{matrix}\right,$$

Реши за b (complex solution)

$$\left\{\begin{matrix}b=a-x^{-\frac{1}{2}}\sqrt{y}\text{; }b=a+x^{-\frac{1}{2}}\sqrt{y}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&y=0\text{ and }x=0\end{matrix}\right,$$

Реши за a

$$\left\{\begin{matrix}a=b-\sqrt{\frac{y}{x}}\text{; }a=b+\sqrt{\frac{y}{x}}\text{, }&y\leq 0\text{ and }x<0\\a=b-\sqrt{\frac{y}{x}}\text{; }a=b+\sqrt{\frac{y}{x}}\text{, }&y\geq 0\text{ and }x>0\\a\in \mathrm{R}\text{, }&y=0\text{ and }x=0\end{matrix}\right,$$

Реши за b

$$\left\{\begin{matrix}b=a+\sqrt{\frac{y}{x}}\text{; }b=a-\sqrt{\frac{y}{x}}\text{, }&y\leq 0\text{ and }x<0\\b=a+\sqrt{\frac{y}{x}}\text{; }b=a-\sqrt{\frac{y}{x}}\text{, }&y\geq 0\text{ and }x>0\\b\in \mathrm{R}\text{, }&y=0\text{ and }x=0\end{matrix}\right,$$