1) \(\left(x-1\right)\left(x+2\right)< 0\Leftrightarrow-2< x< 1\)
vậy \(x=-1;0\)
2) \(\left(x+1\right)\left(2x-4\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-1\end{matrix}\right.\)
vậy \(x=Z\backslash\left\{1;0\right\}\)
3) \(\left(x^2+1\right)\left(x^2-4\right)\le0\)
vì \(x^2+1\ne0\)
\(\Leftrightarrow x^2-4\le0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\le0\Leftrightarrow-2\le x\le2\)
vậy \(x=-2;-1;0;1;2\)
4) \(\left|x\right|\left(x^2-1\right)\ge0\)
ta có \(\left|x\right|\ge0\)
\(\Leftrightarrow x^2-1\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
vậy \(x=Z\backslash\left\{0\right\}\)
1: (x-1)(x+2)<0
=>-2<x<1
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
2: \(\left(x+1\right)\cdot\left(2x-4\right)>=0\)
=>x>=2 hoặc x<=-1
mà x là số nguyên
nên x=Z\{1;0}
3: \(\Leftrightarrow x^2-4< =0\)
=>-2<=x<=2
mà x là số nguyên
nên \(x\in\left\{-2;-1;0;1;2\right\}\)
4: =>(x2-1)>=0
=>x>=1 hoặc x<=-1
=>x=Z\{0}