Question

Помогите пожалуйста!!​

Answer 1

Ответ:

Вот то самое пошаговое решение,думаю помогла

Answer 2

$\displaystyle\bf\\\frac{3}{x^{2} -2x+1} +\frac{2}{x^{2} -1} =-\frac{1}{x+1} \\\\\\\frac{3}{(x-1)^{2} } +\frac{2}{(x-1)(x+1)} +\frac{1}{x+1} =0\\\\\\\frac{3\cdot(x+1)+2\cdot(x-1)+1\cdot(x^{2} -2x+1)}{(x-1)^{2} (x+1)} =0\\\\\\\frac{3x+3+2x-2+x^{2} -2x+1}{(x-1)^{2}(x+1) } =0\\\\\\\frac{x^{2} +3x+2}{(x-1)^{2}(x+1) } =0\\\\\\\left \{ {{x^{2} +3x+2=0} \atop {x-1\neq 0 \ ; \ x+1\neq 0}} \right. \\\\\\\left \{ {{x^{2} +3x+2=0} \atop {x\neq 1 \ ; \ x\neq -1}} \right.\\\\\\x^{2} +3x+2=0$

$\displaystyle\bf\\D=3^{2} -4\cdot 2=9-8=1\\\\\\x_{1} =\frac{-3-1}{2} =\frac{-4}{2} =-2\\\\\\x_{2} =\frac{-3+1}{2} =\frac{-2}{2} =-1 \ - \ neyd\\\\Otvet \ : \ -2$