Решите дробно-рациональное уравнение: 3/(x-4)+2/(x+3)=(x^2+3x-7)/(x^2-x-12).

Ответ:

\[\frac{3}{x - 4} + \frac{2}{x + 3} = \frac{x^{2} + 3x - 7}{x^{2} - x - 12}\]

\[x^{2} - x - 12 = (x + 3)(x - 4)\]

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 12\]

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

\[\frac{3^{\backslash x + 3}}{x - 4} + \frac{2^{\backslash x - 4}}{x + 3} = \frac{x^{2} + 3x - 7}{(x + 3)(x - 4)}\]

\[ОДЗ:x \neq 4;x \neq - 3.\]

\[3x + 9 + 2x - 8 = x^{2} + 3x - 7\]

\[x^{2} + 3x - 5x - 7 - 1 = 0\]

\[x^{2} - 2x - 8 = 0\]

\[D_{1} = 1 + 8 = 9\]

\[x_{1} = 1 + 3 = 4\ (не\ подходит);\]

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

\[Ответ:x = - 2.\]