АЗ. Решить уравнение: 2x-2 x+3 + =5 x+3 x-3

Ответ: x = 4

Решение:

\[\frac{2x-2}{x+3} + \frac{x+3}{x-3} = 5\]

\[\frac{(2x-2)(x-3) + (x+3)(x+3)}{(x+3)(x-3)} = 5\]

\[(2x-2)(x-3) + (x+3)(x+3) = 5(x+3)(x-3)\]

\[2x^2 - 6x - 2x + 6 + x^2 + 6x + 9 = 5(x^2 - 9)\]

\[3x^2 - 2x + 15 = 5x^2 - 45\]

\[2x^2 + 2x - 60 = 0\]

\[x^2 + x - 30 = 0\]

Дискриминант: \[D = 1^2 - 4 \cdot 1 \cdot (-30) = 1 + 120 = 121\]

\[x_1 = \frac{-1 + \sqrt{121}}{2} = \frac{-1 + 11}{2} = \frac{10}{2} = 5\]

\[x_2 = \frac{-1 - \sqrt{121}}{2} = \frac{-1 - 11}{2} = \frac{-12}{2} = -6\]

Проверка:

При x = 5:

\[\frac{2 \cdot 5 - 2}{5+3} + \frac{5+3}{5-3} = \frac{8}{8} + \frac{8}{2} = 1 + 4 = 5\]

При x = -6:

\[\frac{2 \cdot (-6) - 2}{-6+3} + \frac{-6+3}{-6-3} = \frac{-14}{-3} + \frac{-3}{-9} = \frac{14}{3} + \frac{1}{3} = \frac{15}{3} = 5\]

Ответ: x = 5 и x = -6