xét f(x) có nghiệm <=>f(x)=0
<=>x2+2x+1=0
<=>(x+1)2=0
<=>x+1=0
<=>x=-1
Ta có: f(x)=x.x+x+x+1.1=0
=x(x+1)+1(x+1)=0
=(x+1)2=0
=> x+1=0
=> x=-1
Ta có: f(x)=x.x+x+x+1.1=0
=x(x+1)+1(x+1)=0
=(x+1)2=0
=> x+1=0
=> x=-1
\(x^2+2x+1=0\)
\(x.x+2.x+1=0\)
\(x\left(x+2\right)+1=0\)
\(\hept{\begin{cases}x=0\\\left(x+2\right)+1=0\end{cases}}\)
\(\hept{\begin{cases}x=0\\x+2=-1\end{cases}}\)
\(\hept{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy :\(x=\hept{\begin{cases}0\\-3\end{cases}}\)
