1: \(\left|3x-1\right|\ge x^2-x-2\) (1)
TH1: \(x^2-x-2<0\)
=>(1) luôn đúng ∀x thỏa mãn \(x^2-x-2<0\)
=>(x-2)(x+1)<0
=>-1<x<2
TH2: \(x^2-x-2=0\)
=>(x-2)(x+1)=0
=>x=2 hoặc x=-1
Khi x=2 thì |3x-1|=|3*2-1|=|6-1|=5>0
=>Loại
Khi x=-1 thì |3x-1|=|3*(-1)-1|=|-4|=4>0
=>Loại
TH3: \(x^2-x-2>0\)
=>(x-2)(x+1)>0
=>x>2 hoặc x<-1
(1) sẽ tương đương với:
\(\left(3x-1\right)^2\ge\left(x^2-x-2\right)^2\)
=>\(\left(x^2-x-2\right)^2\le\left(3x-1\right)^2\)
=>\(\left(x^2-x-2-3x+1\right)\left(x^2-x-2+3x-1\right)\le0\)
=>\(\left(x^2-4x-1\right)\left(x^2+2x-3\right)\le0\)
TH1: \(\begin{cases}x^2-4x-1\ge0\\ x^2+2x-3\le0\end{cases}\Rightarrow\begin{cases}x^2-4x+4-5\ge0\\ \left(x+3\right)\left(x-1\right)\le0\end{cases}\)
=>\(\begin{cases}\left(x-2\right)^2\ge5\\ -3\le x\le1\end{cases}\Rightarrow\begin{cases}\left[\begin{array}{l}x-2\ge\sqrt5\\ x-2\le-\sqrt5\end{array}\right.\\ -3\le x\le1\end{cases}\)
=>\(\begin{cases}\left[\begin{array}{l}x\ge2+\sqrt5\\ x\le-2+\sqrt5\end{array}\right.\\ -3\le x\le1\end{cases}\)
=>\(-3\le x\le-2+\sqrt5\)
=>\(-3\le x<-1\)
TH2: \(\begin{cases}x^2-4x-1\le0\\ x^2+2x-3\ge0\end{cases}\Rightarrow\begin{cases}x^2-4x+4-5\le0\\ \left(x+3\right)\left(x-1\right)\ge0\end{cases}\)
=>\(\begin{cases}\left(x-2\right)^2-5\le0\\ \left[\begin{array}{l}x\ge1\\ x\le-3\end{array}\right.\end{cases}\Rightarrow\begin{cases}\left(x-2\right)^2\le5\\ \left[\begin{array}{l}x\ge1\\ x\le-3\end{array}\right.\end{cases}\)
=>\(\begin{cases}-\sqrt5\le x-2\le\sqrt5\\ \left[\begin{array}{l}x\ge1\\ x\le-3\end{array}\right.\end{cases}\)
=>\(\begin{cases}-\sqrt5+2\le x\le\sqrt5+2\\ \left[\begin{array}{l}x\ge1\\ x<=-3\end{array}\right.\end{cases}\)
=>\(1\le x\le\sqrt5+2\)
=>\(2
