`|2x-1|=x+4`
`@TH1:2x-1 >= 0=>x >= 1/2 =>|2x-1|=2x-1`
`=>2x-1=x+4`
`=>x=5` (t/m)
`@TH2:2x-1 < 0=>x < 1/2=>|2x-1|=1-2x`
`=>1-2x=x+4`
`=>3x=-3=>x=-1` (t/m)
Vậy `x in {-1;5}`
`|2x-1|=x+4`
=>\(\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-1-x-4=0\\2x-1+x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
TH1: x > 1/2
=> |2x - 1| = 2x - 1
<=> 2x - 1 = x + 4
<=> x = 5 (TM)
TH2: x < 1/2
=> |2x - 1| = 1 - 2x
<=> 1 - 2x = x + 4
<=> -3x = 3
<=> x = -1 (TM)
vậy x ∈ {-1; 5}