Question

Кто решит тот крутой, Срочно!!!
100 балов

Answer 1

$\displaystyle (x^2+3x+1)^2+4(x^2+3x+1)-6=-1\\\\x^2+3x+1=y\\\\y^2+4y-5=0\\\\D=16+40=36\\\\y_{1.2}=\frac{-4 \pm 6}{2}\\\\y_1=-5; y_2=1$

$\displaystyle x^2+3x+1=-5\\\\x^2+3x+6=0\\\\D=9-24<0$

корней нет

$\displaystyle x^2+3x+1=1\\\\x(x+3)=0\\\\x_1=0; x_2=-3$