\(b,\Rightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Rightarrow5\left(x+2\right)=0\\ \Rightarrow x=-2\\ c,\Rightarrow2x\left(x^2-2x+1\right)=0\\ \Rightarrow2x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ d,\Rightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Rightarrow3x\left(-x-2\right)=0\\ \Rightarrow-3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
a)thiếu dấu
b)(x+2)2 -(x+2)(x-3)=0
(x+2)(x+2-x+3)=0
(x+2)5=0
x+2=0
x=-2
c)2x3-4x2+2x=0
2x(x2-2x+1)=0
2x(x-1)2
suy ra 2 trường hợp
x=0
x-1=0=>x=1
d)(x-1)2-(2x+1)2=0
(x-1-2x-1)(x-1+2x+1)=0
(x-2)3x=0
x=0
x=2