{l}{c=b}{X^2=a}{y^2=a^2+b} を解きます| Microsoft 数学ソルバー

$$\left. \begin{array} { l } { c = b } \\ { X ^ { 2 } = a } \\ { y ^ { 2 } = a ^ { 2 } + b } \end{array} \right.$$

c,b,X,a,y を解く (複素数の解)

$$y=-\sqrt{c}\text{, }c\in \mathrm{C}\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=-\sqrt{y^{2}-c}$$

$$y=\sqrt{c}\text{, }c\in \mathrm{C}\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=-\sqrt{y^{2}-c}$$

$$y=-\sqrt{X^{4}+c}\text{, }c\in \mathrm{C}\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{C}\text{, }a=\sqrt{y^{2}-c}$$

$$y=\sqrt{X^{4}+c}\text{, }c\in \mathrm{C}\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{C}\text{, }a=\sqrt{y^{2}-c}$$

c,b,X,a,y を解く

$$\left\{\begin{matrix}\\y=-X^{2}\text{, }c=0\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{R}\text{, }a=|y|\text{; }y=-\sqrt{X^{4}+c}\text{, }c>0\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{R}\text{, }a=\sqrt{y^{2}-c}\text{; }y=0\text{, }c=0\text{, }b=0\text{, }X=0\text{, }a=0\text{; }y=-\sqrt{c}\text{, }c\geq 0\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=\sqrt{y^{2}-c}\text{; }y=0\text{, }c=0\text{, }b=-a^{2}\text{, }X=0\text{, }a=0\text{; }y=X^{2}\text{, }c=0\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{R}\text{, }a=|y|\text{; }y=\sqrt{X^{4}+c}\text{, }c>0\text{, }b=y^{2}-a^{2}\text{, }X\in \mathrm{R}\text{, }a=\sqrt{y^{2}-c}\text{; }y=\sqrt{c}\text{, }c\geq 0\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=\sqrt{y^{2}-c}\text{; }y=\sqrt{c}\text{, }c\geq 0\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=-\sqrt{y^{2}-c}\text{; }y=-\sqrt{c}\text{, }c\geq 0\text{, }b=y^{2}-a^{2}\text{, }X=0\text{, }a=-\sqrt{y^{2}-c}\text{, }&\text{unconditionally}\\y=-\sqrt{X^{4}+c}\text{, }c\in [-X^{4},0)\text{, }b=y^{2}-a^{2}\text{, }X\neq 0\text{, }a=\sqrt{y^{2}-c}\text{; }y=\sqrt{X^{4}+c}\text{, }c\in [-X^{4},0)\text{, }b=y^{2}-a^{2}\text{, }X\neq 0\text{, }a=\sqrt{y^{2}-c}\text{, }&c\leq 0\end{matrix}\right.$$