Решите уравнение методом замены переменной: (x^2+x-3)/2-3/(2x^2+2x-6)=1.

\[\frac{x^{2} + x - 3}{2} - \frac{3}{2x^{2} + 2x - 6} = 1\]

\[\frac{x^{2} + x - 3}{2} - \frac{3}{2 \cdot \left( x^{2} + x - 3 \right)} = 1\]

\[ОДЗ:\ x^{2} + x - 3 \neq 0\]

\[D = 1^{2} - 4 \cdot 1 \cdot ( - 3) =\]

\[= 1 + 12 = 13\]

\[x_{1} \neq \ \frac{- 1 + \sqrt{13}}{2};\ \ \ \]

\[x_{2} \neq \frac{- 1 - \sqrt{13}}{2}.\]

\[Пусть\ x^{2} + x - 3 = t:\]

\[t^{2} - 2t - 3 = 0\]

\[t_{1} + t_{2} = 2\]

\[t_{1} \cdot t_{2} = - 3\]

\[\Longrightarrow t_{1} = 3,\ \ t_{2} = - 1.\]

\[Ответ:\ x = - 3;\ \ x = \pm 2;\ x = 1.\]