Ta có : x3 + 2x2 + 2x + 1 = 0
<=> x3 + 2x2.1 + 2.12.x + 13 = 0
<=> (x + 1)3 = 0
=> x + 1 = 0
=> x = -1
\(x^3+2x^2+2x+1=0\)
\(x^3+x^2+x^2+x+x+1=0\)
\(x^2\left(x+1\right)+x\left(x+1\right)+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)=0\)
Vì \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
x^3 + 2x^2 + 2x +1=0
suy ra:x(x^2 + 2x +1) + 2X =0
suy ra:x=0 hoặc (x+1)^2 =0 hoặc 2x=0
suy ra:x=0 hoặc x = -1 hoặc x=0
vậy x=0 hoặc x =-1
Tao có \(x^3+2x^2+2x+1=0\)
<=>\(x^3+2x^2.1+2.1^2.x+1^3=0\)
<=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
Vậy x=-1