Решите уравнение: (2x^2+3x+1)/((x+1)(x-3))=1.

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

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

\[= (2x + 1)(x + 1);\]

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{( - 3 + 1)}{4} = - \frac{2}{4} = - \frac{1}{2};\]

\[x_{2} = \frac{- 3 - 1}{4} = - 1.\]

\[\frac{(2x + 1)(x + 1)}{(x + 1)(x - 3)} = 1\]

\[ОДЗ:x \neq - 1;\ \ x \neq 3.\]

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

\[2x + 1 = x - 3\]

\[x = - 4.\]

\[Ответ:x = - 4.\]