Avem de rezolvat o ecuatie de gradul 2 reductibila la gradul 1.
[tex]\displaystyle\\
\frac{x^2+x-2}{x-1}=\frac{2x+3}{3}\\\\
Conditii:~~~x-1 \neq 0~~\Rightarrow~~x\neq 1~~\Rightarrow~~x\in \mathbb{R}-\{1\}\\\\
\text{Rezolvare:}\\\\
\frac{x^2+x-2}{x-1}=\frac{2x+3}{3}~~~~unde~~~x=2x-x\\\\
\frac{x^2+2x-x-2}{x-1}=\frac{2x+3}{3}\\\\
\frac{x(x + 2)-(x+2)}{x-1}=\frac{2x+3}{3}\\\\
\frac{(x + 2)(x-1)}{x-1}=\frac{2x+3}{3}~~~\text{Simplificam prima fractie cu } ~~(x-1)\\\\
x+2=\frac{2x+3}{3}\\\\
3(x+2)=2x+3\\
3x+6=2x+3\\
3x-2x=3-6\\
x = \boxed{-3}
[/tex]
