`|2x+2|=|3x-2|`
`<=>` $\left[ \begin{array}{l}2x+2=3x-1\\2x+2=2-3x\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=3\\5x=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=3\\x=0\end{array} \right.$
Vậy `S={0,3}`
`|2x+2|=|3x-2|` `<=>` $\left[ \begin{array}{l}2x+2=3x-2\\2x+2=2-3x\end{array} \right.$ `<=>` $\left[ \begin{array}{l}x=4\\5x=0\end{array} \right.$ `<=>` $\left[ \begin{array}{l}x=4\\x=0\end{array} \right.$ Vậy `S={0,4}`
`|2x+2|=3x-2`
`<=>` $\left[ \begin{array}{l}2x+2=3x-2\\2x+2=2-3x\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=4\\5x=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=4\\x=0\end{array} \right.$
Vậy `S={0,4}`
nếu 3x-2 > 0 (=) x ≥ \(\dfrac{2}{3}\) khi đó | 3x - 2 | = 3x - 2, khi đó ta có pt:
2x + 2 = |3x - 2|
(=) 2x + 2 = 3x - 2
(=) 2x - 3x = -2 - 2
(=) -x = -4
(=) -x . (-1) = -4 . (-1)
(=) x = 4 ( nhận )
nếu 3x-2 < 0 (=) x < \(\dfrac{2}{3}\) khi đó |3x-2 |= -3x + 2, khi đó ta có pt:
2x + 2 = |3x - 2|
(=) 2x + 2 = -3x + 2
(=) 2x + 3x = 2 - 2
(=) 5x = 0
(=) x = 0 ( nhận )
vây S = { 4 ; 0 }