\(x^3-2x^2-x+2>0\)
\(\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)>0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)>0\)
Lại có:\(x-2< x-1< x+1\)
\(\Rightarrow\)Ta xét 2 trường hợp:
TH1:\(\left\{{}\begin{matrix}x-2< 0\\x-1< 0\\x+1>0\end{matrix}\right.\)
\(\Leftrightarrow-1< x< 1\)
TH2:\(\left\{{}\begin{matrix}x-2>0\\x-1>0\\x+1>0\end{matrix}\right.\)
\(\Leftrightarrow x>2\)
\(x^3-2x^2-x+2>0\\ \Leftrightarrow\left(x^3-2x^2\right)-\left(x-2\right)>0\\ \Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)>0\\ \Leftrightarrow\left(x^2-1\right)\left(x-2\right)>0\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-2\right)>0\)
Lập bảng xét dấu:
\(\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)
Vậy bất phương trình có tập nghiệm \(S=\left\{x|-1< x< 1;x>2\right\}\)
