Вопрос в картинках...

Пусть $x^2+2x+1=(x+1)^2=t$
$\frac{t}{t+1}+ \frac{t+1}{t+2} = \frac{7}{6} \\ \frac{t(t+2)+(t+1)^2}{(t+1)(t+2)}= \frac{7}{6} \\ \frac{t^2+2t+t^2+2t+1}{t^2+3t+2} = \frac{7}{6} \\ \left \{ {{t \neq -2;-1} \atop {7t^2+21t+14=12t^2+24t+6}} \right. \\ \left \{ {{t \neq -2;-1} \atop {5t^2+3t-8=0}} \right. \\ \left \{ {{t \neq -2;-1} \atop {t=-1.6;1 }} \right. \\ t=-1.6;1$
$(x+1)^2=-1.6 \\ x\in\varnothing$ или $(x+1)^2=1\\ \left [ {{x+1=1} \atop {x+1=-1}} \right. \\ \left [ {{x=0} \atop {x=-2}} \right.$

Ответ: x = -2; 0