{l}{ax^{2}+b^2-2ab^2=(b-ab)x}{6x^2+(b-2a)x=b^2+ab} шешу

$$\left. \begin{array} { l } { a x ^ { 2 } + b ^ { 2 } - 2 a b ^ { 2 } = ( b - a b ) x } \\ { 6 x ^ { 2 } + ( b - 2 a ) x = b ^ { 2 } + a b } \end{array} \right.$$

x мәнін табыңыз (complex solution)

$$\left\{\begin{matrix}x=b\text{, }&b=0\text{ or }a=2b\\x=-2b+\frac{b}{a}\text{, }&a\neq 0\text{ and }\left(b=0\text{ or }a=\frac{2}{3}\text{ or }a=\frac{\sqrt{49b^{2}+12b}-7b}{2}\text{ or }a=\frac{-\sqrt{49b^{2}+12b}-7b}{2}\right)\end{matrix}\right.$$

x мәнін табыңыз

$$\left\{\begin{matrix}x=b\text{, }&b=0\text{ or }a=2b\\x=-2b+\frac{b}{a}\text{, }&\left(a\neq 0\text{ and }b=0\right)\text{ or }a=\frac{2}{3}\text{ or }\left(a=\frac{\sqrt{49b^{2}+12b}-7b}{2}\text{ and }b>0\right)\text{ or }\left(a=\frac{\sqrt{49b^{2}+12b}-7b}{2}\text{ and }b\leq -\frac{12}{49}\right)\text{ or }\left(a=\frac{-\sqrt{49b^{2}+12b}-7b}{2}\text{ and }b>0\right)\text{ or }\left(a=\frac{-\sqrt{49b^{2}+12b}-7b}{2}\text{ and }b\leq -\frac{12}{49}\right)\end{matrix}\right.$$