`|2x+1|-|5x-2|=3`
Nếu `x>=2/5=>|2x+1|=2x+1,|5x-2|=5x-2`
`pt<=>2x+1-(5x-2)=3`
`<=>2x+1-5x+2=3`
`<=>-3x+3=3`
`<=>-3x=0`
`<=>x=0(l)`
Tương tự:
Với `x<=-1/2`
`pt<=>-(2x+1)+5x-2=3`
`<=>3x-1=3`
`<=>3x=4<=>x=4/3(l)`
`-1/2<=x<=5/2`
`pt<=>2x+1+5x-2=3`
`<=>7x-1=3`
`<=>x=4/7(tm)`
Vậy `S={4/7}`
TH1: \(x\ge\dfrac{2}{5}\)
=> \(\left\{{}\begin{matrix}\left|2x+1\right|=2x+1\\\left|5x-2\right|=5x-2\end{matrix}\right.\)
PT <=> 2x + 1 - 5x + 2 = 3
<=> -3x = 0
<=> x = 0 (loại)
TH2: \(\dfrac{-1}{2}\le x< \dfrac{2}{5}\)
<=> \(\left\{{}\begin{matrix}\left|2x+1\right|=2x+1\\\left|5x-2\right|=2-5x\end{matrix}\right.\)
PT <=> 2x + 1 + 5x -2 = 3
<=> 7x = 4
<=> x = \(\dfrac{4}{7}\) (loại)
TH3: \(x< \dfrac{-1}{2}\)
=> \(\left\{{}\begin{matrix}\left|2x+1\right|=-2x-1\\\left|5x-2\right|=2-5x\end{matrix}\right.\)
PT <=> -2x - 1 + 5x - 2 = 3
<=> 3x = 6
<=> x = 2 (loại)
KL: x \(\in\varnothing\)
