{l}{x=frac{x_{1}x_{2}+y_{1}y_{2}}{x_2^2+y_2^2}}{y={(x_2^2+y_2^2)}}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a} を解きます| Microsoft 数学ソルバー

$$\left. \begin{array} { l } { x = \frac{x_{1} x_{2} + y_{1} y_{2}}{x_{2} ^ {2} + y_{2} ^ {2}} }\\ { y = {(x_{2} ^ {2} + y_{2} ^ {2})} }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

x,y,z を解く (複素数の解)

$$x=\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}$$

$$y=x_{2}^{2}+y_{2}^{2}$$

$$z=x_{2}^{2}+y_{2}^{2}$$

$$\left(x_{2}=-\sqrt{b-y_{2}^{2}}\text{ and }arg(iy_{2})\geq \pi \text{ and }y_{2}\neq 0\text{ and }arg(-iy_{2})\geq \pi \text{ and }a=b\right)\text{ or }\left(x_{2}=-\sqrt{b-y_{2}^{2}}\text{ and }b\neq 0\text{ and }a=b\right)\text{ or }\left(x_{2}=\sqrt{b-y_{2}^{2}}\text{ and }arg(iy_{2})\geq \pi \text{ and }y_{2}\neq 0\text{ and }arg(-iy_{2})\geq \pi \text{ and }a=b\right)\text{ or }\left(x_{2}=\sqrt{b-y_{2}^{2}}\text{ and }b\neq 0\text{ and }a=b\right)$$

x,y,z を解く

$$x=\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}$$

$$y=x_{2}^{2}+y_{2}^{2}$$

$$z=x_{2}^{2}+y_{2}^{2}$$

$$a=b\text{ and }\left(y_{2}\neq 0\text{ or }x_{2}\neq 0\right)\text{ and }\left(b>y_{2}^{2}\text{ or }y_{2}\neq 0\right)\text{ and }b\geq y_{2}^{2}\text{ and }|x_{2}|=\sqrt{b-y_{2}^{2}}$$