Решите уравнение

Решите задачу:

$\displaystyle\frac{3}{x+1} +1= \frac{10}{x^2+2x+1} \\ \\ ODZ:X \neq 1 \\ \\ \frac{3}{x+1} +1 - \frac{10}{x^2+2x+1} =0 \\ \\ \frac{3}{x+1} +1 - \frac{10}{(x+1)^2} =0 \\ \\ \frac{3(x+1)+(x+1)^2-10}{(x+1)^2} =0 \\ \\ \frac{3x+3+(x+1)^2-10}{(x+1)^2} =0 \\ \\ \frac{3x-7+x^2+2x+1}{(x+1)^2} =0 \\ \\ \frac{5x-6+x^2}{(x+1)^2} =0 \\ \\$

$\displaystyle5x-6+x^2=0 \\ \\ x^2+5x-6=0 \\ \\ D=25+24=49=7^2 \\ \\ x_{1} = \frac{-5+7}{2} = 1 \\ \\ x_2= \frac{-5-7}{2} =-6 \\ \\ Otvet:x_1=1 , x_2=-6$