Bài này giải như sau :
ta có : x2+3x+2=0
⇔ x2 + 2x + x + 2
⇔( x2 + x) + (2x+2)
⇔x ( x + 1 ) + 2 ( x + 1 )
⇔( x + 1)+( x + 2 ) = 0 ⇒ x = (-1) hay (-2)
\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)=0\)
Đến đây bí :(
\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-1;-2\right\}\)
Z