\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(2\left(x+3\right)+x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
<=> (x+3)(x+2)=0
TH1 x+3=0 <=> x=-3
TH2 x+2=0 <=> x=-2
Vậy....
\(b,\Rightarrow\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(x+2\right)=0\\ \Rightarrow\left(2x-3-x-2\right)^2=0\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)
Lời giải:
a. $2(x+3)+x(3+x)=0$
$\Leftrightarrow (x+3)(2+x)=0$
$\Leftrightarrow x+3=0$ hoặc $2+x=0$
$\Leftrightarrow x=-3$ hoặc $x=-2$
b.
$(2x-3)^2-(4x-6)(x+2)+x^2+4x+4=0$
$\Leftrightarrow (2x-3)^2-2(2x-3)(x+2)+(x+2)^2=0$
$\Leftrightarrow [(2x-3)-(x+2)]^2=0$
$\Leftrightarrow (x-5)^2=0$
$\Leftrightarrow x=5$
