\(\left(x+1\right).\left(x^2-4\right)=0\)
\(\left(x+1\right).\left(x^2-2^2\right)=0\)
\(\left(x+1\right).\left(x-2\right).\left(x+2\right)\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x-2=0\\x+2=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0-1\\x=0+2\\x=0-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\x=2\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{1;2;-2\right\}\)
Ta có
\(\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\\x+2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=-2\end{array}\right.\)
Vậy x=1 ; x=2 ; x= - 2
