Đáp án:
Giải thích các bước giải:
`4)`
`|2 1/2 + x| - (-2)/3 = 3`
`-> |5/2+x| + 2/3 = 3`
`-> |5/2+x| = 3 - 2/3`
`-> |5/2+x| = 7/3`
TH1 :
`5/2 + x = 7/3`
`-> x = 7/3 - 5/2`
`-> x = -1/6`
TH2 :
`5/2 + x = -7/3`
`-> x = -7/3 - 5/2`
`-> x = -29/6`
Vậy `x \in {-1/6,-29/6}`
`5)`
`-5/7-|1/2-x|=-11/4`
`-> |1/2-x| = -5/7 - (-11/4)`
`-> |1/2-x| =57/28`
TH1 :
`1/2 - x = 57/28`
`-> x = 1/2 - 57/28`
`-> x = -43/28`
TH2 :
`1/2 - x = -57/28`
`-> x = 1/2 - (-57/28)`
`-> x = 71/28`
Vậy `x \in {-43/28,71/28}`
`6)`
`|x-1/3|+4/5=|(-3,2)+2/5|`
`-> |x-1/3| = |-16/5+2/5|-4/5`
`-> |x-1/3| = 14/5 - 4/5`
`-> |x-1/3| = 10/5`
`-> |x-1/3| = 2`
TH1 :
`x - 1/3 = 2`
`-> x = 2 + 1/3`
`-> x = 7/3`
TH2 :
`x - 1/3 = -2`
`-> x = -2 + 1/3`
`-> x = -5/3`
Vậy `x \in {7/3,-5/3}`
