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|x-1|=|2|+|-6| 3/4+ |x-5/2| = 1/4 3/2-|x-2/3| = 4/3 |x-2|= |-1/2| + |-3/4|
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Philomena
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Đáp án:
Giải thích các bước giải:
`|x- 1|= |2|+ |-6|`
`|x+ 1|= -2+ 6`
`|x+ 1|= 4`
`⇒ |x+ 1|= ±4`
`Th1: x+ 1= -4`
`x= -4- 1`
`x= -5`
`Th2: x+ 1= 4`
`x= 4- 1`
`x= 3`
Vậy `x∈ \text{{-5; 3}}`
`3/4+ |x- 5/2|= 1/4`
`|x- 5/2|= 1/4- 3/4`
`|x- 5/2|= -1/2`
`x- 5/2= -1/2`
`x= -1/2+ 5/2`
`x= 4/2= 2`
`3/2- |x- 2/3|= 4/3`
`|x- 2/3|= 4/3+ 3/2`
`|x- 2/3|= 8/6+ 9/6`
`|x- 2/3|= 17/6`
`|x+ 2/3|= ±17/6`
`Th1: x+ 2/3= 17/6`
`x= 17/6- 2/3`
`x= 17/6- 4/6`
`x= 13/6`
`Th2: x+ 2/3= -17/6`
`x= -17/6- 2/3`
`x= -17/6+ 4/6`
`x= -13/6`
Vậy `x∈ \text{{-13/6; 13/6}}`
`|x- 2|= |-1/2|+ |-3/4|`
`|x- 2|= 1/2+ 3/4`
`|x- 2|= 2/4+ 3/4`
`|x- 2|= 5/4`
`⇒ |x- 2|= ±5/4`
`Th1: x- 2= 5/4`
`x= 5/4+ 2`
`x= 5/4+ 8/4`
`x= 13/4`
`Th2: x- 2= -5/4`
`x= -5/4+ 2`
`x= -5/4+ 8/4`
`x= 3/4`
Vậy `x∈ \text{{3/4; 13/4}}`
| x-1 | = |2| + | -6 |
| x-1 | = 2+ 6
| x-1 | = 8
=> x-1 = 8 hoặc x -1 = -8
x = 8 + 1 x = (-8 ) + 1
x = 9 x = ( -7 )
vậy x ∈ { 9 ; -7 }
3/4 + | x – 5/2 | = 1/4
| x – 5/2 | = 1/4 – 3/4
| x – 5/2 | = -1/2
vì | x – 5/2 | ≥ 0 mà -1/2 < 0
=> x ∈ ∅
Vậy x ∈ ∅
3/2 – | x -2/3 | = 4/3
| x – 2/3 | = 3/2 – 4/3
| x – 2/3 | = 1/6
=> x – 2/3 = 1/6 hoặc x – 2/3 = -1/6
x = 1/6 + 2/3 x = -1/6 + 2/3
x = 5/6 x = 1/2
vậy x ∈ { 5/6 ; 1/2}
| x – 2 | = | -1/2 | + | -3/4 |
| x – 2 | = 1/2 + 3/4
| x- 2 | = 5/4
=> x – 2 = 5/4 hoặc x – 2 = -5/4
x = 5/4 + 2 x = -5/4 + 2
x = 13/4 x = 3/4
vậy x ∈ { 3/4 ; 13/4 }