Question

(x^2+2x+1)(x^2+2x)=12 решите на листочке

Answer 1

$(x^2+2x+1)(x^2+2x)=12\\\\ x^2+2x+1=(x+1)^2=t \geq 0\\\\ t*(t-1)=12\\\\ t^2-t-12=0\\\\ t^2-4t+3t-12=0\\\\ t(t-4)+3(t-4)=0\\\\ (t+3)(t-4)=0\\\\ t=-3\ \ or\ \ t=4\\\\ ----------\\\\ (x+1)^2=4\\\\ (x+1)^2-2^2=0\\\\ ([x+1]-2)*([x+1]+2)=0\\\\ (x-1)*(x+3)=0\\\\ x-1=0\ \ or\ \ x+3=0\\\\ x=1\ \ or\ \ x=-3$

ответ: $-3;\ 1$

Answer 2

Решение на листке.
ответ: {-1; 4}