Giải (a^2-b^2)/a+b-(a+b)^2+2(a^2+ab)=1

$$\frac { ( a ^ { 2 } - b ^ { 2 } ) } { a + b } - ( a + b ) ^ { 2 } + 2 ( a ^ { 2 } + a b ) = 1$$

Tìm b (complex solution)

$$b=\frac{-\sqrt{4a^{2}+4a-3}-1}{2}$$

$$b=\frac{\sqrt{4a^{2}+4a-3}-1}{2}\text{, }a\neq \frac{1}{2}$$

Tìm a

$$\left\{\begin{matrix}\\a=\frac{-\sqrt{4b^{2}+4b+5}-1}{2}\text{, }&\text{unconditionally}\\a=\frac{\sqrt{4b^{2}+4b+5}-1}{2}\text{, }&b\neq -\frac{1}{2}\end{matrix}\right,$$

Tìm b

$$b=\frac{-\sqrt{4a^{2}+4a-3}-1}{2}$$

$$b=\frac{\sqrt{4a^{2}+4a-3}-1}{2}\text{, }a\leq -\frac{3}{2}\text{ or }a>\frac{1}{2}$$