Լուծեք {l}{(x+2)^2+(y-1)^2=1}{z=my}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { {(x + 2)} ^ {2} + {(y - 1)} ^ {2} = 1 }\\ { z = m y }\\ { a = z }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Լուծել x, y, z-ի համար (complex solution)

$$\left\{\begin{matrix}x=\frac{i\sqrt{d}\sqrt{d-2m}-2m}{m}\text{, }y=\frac{d}{m}\text{, }z=d\text{; }x=\frac{-i\sqrt{d}\sqrt{d-2m}-2m}{m}\text{, }y=\frac{d}{m}\text{, }z=d\text{, }&m\neq 0\text{ and }c=d\text{ and }a=d\text{ and }b=d\\x\in \mathrm{C}\text{, }y=\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{; }x\in \mathrm{C}\text{, }y=-\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{, }&a=0\text{ and }m=0\text{ and }b=0\text{ and }c=0\text{ and }d=0\end{matrix}\right.$$

Լուծել x, y, z-ի համար

$$\left\{\begin{matrix}x=-\frac{\sqrt{c\left(2m-c\right)}+2|m|}{|m|}\text{, }y=\frac{c}{m}\text{, }z=c\text{; }x=\frac{\sqrt{c\left(2m-c\right)}-2|m|}{|m|}\text{, }y=\frac{c}{m}\text{, }z=c\text{, }&\left(c<0\text{ or }m\geq \frac{c}{2}\right)\text{ and }\left(m\geq 0\text{ or }m\geq \frac{c}{2}\text{ or }c<0\right)\text{ and }\left(c>0\text{ or }m\leq \frac{c}{2}\right)\text{ and }c\neq 0\text{ and }d=c\text{ and }a=c\text{ and }b=c\\x\in \begin{bmatrix}-3,-1\end{bmatrix}\text{, }y=\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{; }x\in \begin{bmatrix}-3,-1\end{bmatrix}\text{, }y=-\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{, }&a=0\text{ and }m=0\text{ and }b=0\text{ and }c=0\text{ and }d=0\end{matrix}\right.$$