\(x^2+5x+6=0\)
\(x^2+3x+2x+6=0\)
\(x\left(x+3\right)+2\left(x+3\right)=0\)
\(\left(x+3\right)\left(2+x\right)=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
hoặc \(2+x=0\Rightarrow x=-2\)
\(x^2+5x+6=0\\ \Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x^2+2x\right)+\left(3x+6\right)=0\\ \Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy \(x=-2\) hoặc \(x=-3\)
x2 + 5x + 6 = 0
=> x2 + 3x + 2x + 6 = 0
=> (x2 + 3x) + (2x + 6) = 0
=> x(x + 3) + 2(x + 3) = 0
=> (x + 3)(x + 2) = 0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)
