解決left(x-3right)^2+left(y+7right)^2+left(z-6right)^2=8 |Microsoft數學求解器

$${ \left(x-3 \right) }^{ 2 } + { \left(y+7 \right) }^{ 2 } + { \left(z-6 \right) }^{ 2 } =8$$

解 x (復數求解)

$$x=-\sqrt{-y^{2}-14y-z^{2}+12z-77}+3$$

$$x=\sqrt{-y^{2}-14y-z^{2}+12z-77}+3$$

解 y (復數求解)

$$y=-\left(\sqrt{-x^{2}+6x-z^{2}+12z-37}+7\right)$$

$$y=\sqrt{-x^{2}+6x-z^{2}+12z-37}-7$$

解 x

$$\left\{\begin{matrix}x=\sqrt{-z^{2}+12z-28}+3\text{; }x=-\sqrt{-z^{2}+12z-28}+3\text{, }&z\geq 6-2\sqrt{2}\text{ and }z\leq 2\sqrt{2}+6\text{ and }y=-7\\x=-\sqrt{-y^{2}-14y-z^{2}+12z-77}+3\text{; }x=\sqrt{-y^{2}-14y-z^{2}+12z-77}+3\text{, }&y\leq \sqrt{-z^{2}+12z-28}-7\text{ and }y\geq -\sqrt{-z^{2}+12z-28}-7\text{ and }z>6-2\sqrt{2}\text{ and }z<2\sqrt{2}+6\end{matrix}\right.$$

解 y

$$\left\{\begin{matrix}y=\sqrt{-z^{2}+12z-28}-7\text{; }y=-\sqrt{-z^{2}+12z-28}-7\text{, }&z\geq 6-2\sqrt{2}\text{ and }z\leq 2\sqrt{2}+6\text{ and }x=3\\y=-\sqrt{-x^{2}+6x-z^{2}+12z-37}-7\text{; }y=\sqrt{-x^{2}+6x-z^{2}+12z-37}-7\text{, }&x\leq \sqrt{-z^{2}+12z-28}+3\text{ and }x\geq -\sqrt{-z^{2}+12z-28}+3\text{ and }z>6-2\sqrt{2}\text{ and }z<2\sqrt{2}+6\end{matrix}\right.$$