`e,`
`(x + 3/5) . (3/2 - 2x) = 0`
`=>` Xảy ra `1` trong `2` trường hợp sau:
`+) x + 3/5 = 0`
`=> x = 0 - 3/5`
`=> x = (-3)/5`
`+) 3/2 - 2x = 0`
`=> 2x = 3/2 - 0`
`=> 2x = 3/2`
`=> x = 3/2 : 2`
`=> x = 3/4`
Vậy `x ∈ {(-3)/5 ; 3/4}`
`f,`
`((-5)/4 x + 1/2) . [(3/5 - (-2/3x)] = 0`
`=>` Xảy ra `1` trong `2` trường hợp sau:
`+) (-5)/4 x + 1/2 = 0`
`=> (-5)/4 x = 0 - 1/2`
`=> (-5)/4 x = (-1)/2`
`=> x = (-1)/2 : (-5)/4`
`=> x = 2/5`
`+) 3/5 - (-2/3 x) = 0`
`=> 3/5 + 2/3 x = 0`
`=> 2/3 x = 0 - 3/5`
`=> 2/3 x = -3/5`
`=> x = (-3)/5 : 2/3`
`=> x = (-9)/10`
Vậy `x ∈ {2/5 ; (-9)/10}`
