`|x^2-2x-3|=|2x-5|+1`
`<=>|(x+1)(x-3)|=|2x-5|+1`
Nếu `x>=3=>|x^2-2x-3|=x^2-2x-3,|2x-5|=2x-5`
`pt<=>x^2-2x-3=2x-5+1`
`<=>x^2-2x-3=2x-4`
`<=>x^2-4x+1=0`
`<=>(x-2)^2=3`
`<=>x=sqrt3+2(do\ x>=3)`
Nếu `x<=-1=>|x^2-2x-3|=3+2x-x^2,|2x-5|=5-2x`
`pt<=>3+2x-x^2=5-2x+1`
`<=>3+2x-x^2=6-2x`
`<=>x^2-4x+3=0`
`<=>(x-1)(x-3)=0`
Vì `x<=-1=>x-1<0,x-3<0`
`=>` pt vô nghiệm
Vậy `x=sqrt3+2`
\(\left|x^2-2x-3\right|=\left|2x-5\right|+1\) (1)
\(\Leftrightarrow\left|\left(x-3\right)\left(x+1\right)\right|=\left|2x-5\right|+1\)
BXD:
TH1: \(x\le-1\)
PT (1) \(\Leftrightarrow x^2-2x-3=-\left(2x-5\right)+1\) \(\Leftrightarrow x^2-9=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(L\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
TH2:\(-1< x< \dfrac{5}{2}\)
PT (1)\(\Leftrightarrow-\left(x^2-2x-3\right)=-\left(2x-5\right)+1\) \(\Leftrightarrow-x^2+4x-3=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(L\right)\\x=1\left(tm\right)\end{matrix}\right.\)
TH3: \(\dfrac{5}{2}\le x\le3\)
PT (1) \(\Leftrightarrow-\left(x^2-2x-3\right)=\left(2x-5\right)+1\)\(\Leftrightarrow-x^2+7=0\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\left(tm\right)\\x=-\sqrt{7}\left(L\right)\end{matrix}\right.\)
TH4:\(x>3\)
PT (1)\(\Leftrightarrow x^2-2x-3=2x-5+1\)\(\Leftrightarrow x^2-4x+1=0\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\left(tm\right)\\x=2-\sqrt{3}\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-3;1;\sqrt{7};2+\sqrt{3}\right\}\)
