Tìm X A : ( 5x – 1 ) × ( 2x -1/3 ) =0 B : ( x -1 ) ( x +2) C : (x – 3 ) ( x + 2) bé hơn 0 17/07/2021 Bởi Margaret Tìm X A : ( 5x – 1 ) × ( 2x -1/3 ) =0 B : ( x -1 ) ( x +2) C : (x – 3 ) ( x + 2) bé hơn 0
A: `(5x-1).(2x-1/3) =0` Th1: `5x-1 =0 <=> x= 1/5` Th2 : `2x-1/3 = 0<=> x = 1/6` B: `(x-1)(x+2) > 0` Th1 : `x-1 > 0 ; x +2> 0 <=> x > 1 ; x > -2 <=> x >1` Th2 : `x-1 < 0 ; x +2< 0 <=> x < 1 ; x < -2 <=> x <-2` Vậy `x <-2 ; x > 1` C : `(x-3)(x+2) < 0` Th1: `x-3 > 0;x +2 < 0<=>x > 3 ; x < -2 ` Th2 : `x-3 <0 ; x + 2>0 <=> x < 3 ; x > -2 <=> -2 < x < 3` `S = \mathbbR \\ {-2;3}` Bình luận
A: `(5x-1).(2x-1/3) =0`
Th1: `5x-1 =0 <=> x= 1/5`
Th2 : `2x-1/3 = 0<=> x = 1/6`
B: `(x-1)(x+2) > 0`
Th1 : `x-1 > 0 ; x +2> 0 <=> x > 1 ; x > -2 <=> x >1`
Th2 : `x-1 < 0 ; x +2< 0 <=> x < 1 ; x < -2 <=> x <-2`
Vậy `x <-2 ; x > 1`
C : `(x-3)(x+2) < 0`
Th1: `x-3 > 0;x +2 < 0<=>x > 3 ; x < -2 `
Th2 : `x-3 <0 ; x + 2>0 <=> x < 3 ; x > -2 <=> -2 < x < 3`
`S = \mathbbR \\ {-2;3}`