+bx=0-x(a^2+b^2)-b(ax-1) を解きます| Microsoft 数学ソルバー

$$+ b x = 0 - x ( a ^ { 2 } + b ^ { 2 } ) - b ( a x - 1 )$$

a を解く (複素数の解)

$$\left\{\begin{matrix}a=\frac{-bx+\sqrt{bx\left(4-4x-3bx\right)}}{2x}\text{; }a=-\frac{bx+\sqrt{bx\left(4-4x-3bx\right)}}{2x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&x=0\text{ and }b=0\end{matrix}\right.$$

b を解く (複素数の解)

$$\left\{\begin{matrix}b=\frac{\sqrt{\left(3ax+x-1\right)\left(-ax+x-1\right)}-ax-x+1}{2x}\text{; }b=\frac{-\sqrt{\left(3ax+x-1\right)\left(-ax+x-1\right)}-ax-x+1}{2x}\text{, }&x\neq 0\\b=0\text{, }&x=0\end{matrix}\right.$$

a を解く

$$\left\{\begin{matrix}a=\frac{-bx+\sqrt{bx\left(4-4x-3bx\right)}}{2x}\text{; }a=\frac{-bx-\sqrt{bx\left(4-4x-3bx\right)}}{2x}\text{, }&\left(b\geq -\frac{4}{3}+\frac{4}{3x}\text{ or }b\geq 0\right)\text{ and }\left(b\leq 0\text{ or }b\leq -\frac{4}{3}+\frac{4}{3x}\right)\text{ and }x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }b=0\end{matrix}\right.$$