Алгебра, 8 класс. Решите уравнение: [latex] \frac{3}{x-1} - \frac{x}{x+1} = \frac{2}{1-x^2} [/latex]

[latex] \frac{3}{x-1}-\frac{x}{x+1}=-\frac{2}{(x-1)(x+1)} |(x+1)(x-1) \\ 3(x+1)-x(x-1)+2=0 \\ 3x+3-x^2+x+2=0 \\ -x^2+4x+5=0 \\ x^2-4x-5=0 \\ D=16+4*5=36 \\ x_{1,2} = \frac{4\pm6}{2} \\ x_1 = \frac{4+6}{2} = \frac{10}{2} = 5 \\ x_2 = \frac{4-6}{2} = \frac{-2}{2} = -1 \\ x \neq \pm1, x \neq x_2  \\ x \in \{5\} [/latex]