Решите уравнение: 1/(x-3)-(x+8)/(2x^2-18)=1/(3-x)-1.

Ответ:

\[\frac{1}{x - 3} - \frac{x + 8}{2x^{2} - 18} = \frac{1}{3 - x} - 1\]

\[\frac{1}{x - 3} - \frac{x + 8}{2\left( x^{2} - 9 \right)} - \frac{1}{3 - x} + 1 = 0\]

\[\frac{1}{x - 3} - \frac{x + 8}{2(x - 3)(x + 3)} + \frac{1}{x - 3} + 1 = 0\]

\[\frac{2^{\backslash 2(x + 3)}}{x - 3} - \frac{x + 8}{2(x - 3)(x + 3)} + 1^{\backslash 2x^{2} - 18} = 0\]

\[\frac{4x + 12 - x - 8 + 2x^{2} - 18}{2(x - 3)(x + 3)} = 0\]

\[\frac{2x^{2} + 3x - 14}{2(x - 3)(x + 3)} = 0\]

\[ОДЗ:x \neq \pm 3.\]

\[2x^{2} + 3x - 14 = 0\]

\[D = 9 + 112 = 121\]

\[x_{1} = \frac{- 3 + 11}{4} = 2;\]

\[x_{2} = \frac{- 3 - 11}{4} = - \frac{14}{4} = - 3,5.\]

\[Ответ:x = - 3,5;x = 2.\]