Question

Помогите пожалуйста с одним примером

Answer 1

$x\neq1 \: \: \: \: \: x\neq - 1 \: \: \: \: \: x\neq0 \\ \frac{3x - 5}{ {x}^{2} - 1} = \frac{3x + 2}{ {x}^{2} + x } - \frac{6x - 5}{ {x}^{2} - x } \\ \frac{3x - 5}{(x - 1)(x + 1)} - \frac{3x + 2}{x(x + 1)} + \frac{6x - 5}{x(x - 1)} = 0 \\ \frac{x(3x - 5) - (3x + 2)(x - 1) + (6x - 5)(x + 1)}{x(x - 1)(x + 1)} = 0 \\ \frac{3x {}^{2} - 5x - 3 {x}^{2} + 3x - 2x + 2 + 6 {x}^{2} + 6x - 5x - 5}{x(x - 1)(x + 1)} = 0 \\ \frac{6{x}^{2} - 3x - 3}{x(x - 1)(x + 1)} = 0 \\ 6 {x}^{2} - 3x - 3 = 0 \\ 2 {x}^{2} - x - 1 = 0 \\ a = 2 \\ b = -1 \\ c = - 1\\ D = {b}^{2} - 4ac = ( -1 ) {}^{2} - 4 \times 2 \times ( - 1) = \\ = 1 + 8 = 9 \\ x_{1} = \frac{1 - 3}{2 \times 2} = - \frac{2}{4} = - 0.5 \\ x_{2} = \frac{1 + 3}{ 2\times 2} = \frac{4}{4} = 1$

Второй корень не подходит

Ответ: х = - 0,5