Найдите корни уравнения: (x-7)/(x-2)+(x+4)/(x+2)=1.

Ответ:

\[\frac{x - 7}{x - 2} + \frac{x + 4}{x + 2} = 1\]

\[ОДЗ:\ \ x - 2 \neq 0;\ \ x \neq 2\]

\[\ \ \ \ \ \ \ x + 2 \neq 0;\ \ x \neq - 2\]

\[\frac{(x - 7)(x + 2) + (x + 4)(x - 2)}{(x - 2)(x + 2)} = 1\]

\[2x^{2} - 3x - 22 - x^{2} + 4 = 0\]

\[x^{2} - 3x - 18 = 0\]

\[D = b^{2} - 4ac =\]

\[= 9 - 4 \cdot 1 \cdot ( - 18) = 9 + 72 =\]

\[= 81\]

\[x_{1} = \frac{3 + 9}{2} = \frac{12}{2} = 6\]

\[x_{2} = \frac{3 - 9}{2} = - \frac{6}{2} = - 3\]

\[Ответ:x = 6\ \ и\ \ x = - 3.\]