\(a,\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-4\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
a, ( x + 1 )2 - x - 1 = 0
( x + 1 )2 - ( x + 1 ) = 0
( x + 1 ) ( x + 1 - 1 ) = 0
=> x ( x + 1 ) = 0
TH1: x = 0 TH2: x + 1 = 0
=> x = -1
Vậy x = 0 hoặc x = -1
b, x3 - 4x = 0
x ( x2 - 4 ) = 0
=> x ( x - 2 ) ( x + 2 ) = 0
TH1: x = 0 TH2: x - 2 = 0 TH3: x + 2 = 0
=> x = 2 => x = -2
Vậy x = 0 hoặc x = 2 hoặc x = -2
