`a)`
`|3x - 5| + |2x + 3| = 7`
Xét: `x > 5/3` Ta có: `3x - 5 + 2x + 3 = 7`
`<=> 5x - 2 = 7 <=> 5x = 9 => x = 9/5` (thỏa mãn)
Xét `\frac{-2}{3} ≤ x ≤ 5/3` Ta có: `- (3x - 5) + 2x + 3 = 7`
`<=> - 3x + 5 + 2x + 3 = 7 <=> - x + 8 = 7 => x = 1` (Thỏa mãn)
Xét `x < \frac{-2}{3}` Ta có: `- 3x + 5 - 2x - 3 = 7`
`<=> - 5x + 2 = 7 <=> - 5x = 5 => x = - 1` (loại)
Vậy `x = { 9/5; 1 }`
`b)`
`x | + | x + 2 | = 3`
`=> 2x = 5`
`=> x=5/2`
Vậy `x = { 5/2 }`
`c)`
`|3x - 5| = |x + 2|`
`=> 3x - 5 = x + 2` hoặc `3x - 5 = - (x + 2)`
TH1: Nếu `3x - 5 = x + 2`
`=> 3x - x = 2 + 5`
`=> 2x = 7`
`=> x = 7/2`
TH2: Nếu `3x - 5 = - (x + 2)`
`=> 3x - 5 = -x - 2`
`=> 3x + x = -2 + 5`
`=> 4x = 3`
`=> x = 3/4`
Vậy `x = { 7/2 ; x = 3/4 }`