Giải thích các bước giải:
`|x-3/2|+|3/2-x|=2`
Ta có:
`(x-3/2)+(3/2-x)`
`=x-3/2+3/2-x`
`=0`
`=>(x-3/2);(3/2-x)`là 2 số đối nhau
`=>|x-3/2|=|3/2-x|`
`=>|x-3/2|+|x-3/2|=2`
`=>2|x-3/2|=2`
`=>|x-3/2|=2:2`
`=>|x-3/2|=1`
`=>``{(x-3/2=1),(x-3/2=-1):}`
`=>``{(x=1+3/2),(x=-1+3/2):}`
`=>``{(x=2/2+3/2),(x=-2/2+3/2):}`
`=>``=>``{(x=5/2),(x=1/2):}`
Vậy `x in{5/2;1/2}`
`|11/4- [|x-3/2|] |=2`
`=>``{(11/4- |x-3/2|=2),(11/4- |x-3/2|=-2):}`
`=>``{(|x-3/2|=11/4-2),(|x-3/2|=11/4+2):}`
`=>``{(|x-3/2|=3/4),(|x-3/2|=19/4):}`
`=>``{(x-3/2=3/4),(x-3/2=-3/4),(x-3/2=19/4),(x-3/2=-19/4):}`
`=>``{(x=3/4+3/2),(x=-3/4+3/2),(x=19/4+3/2),(x=-19/4+3/2):}`
`=>``{(x=9/4),(x=3/4),(x=25/4),(x=-13/4):}`
Vậy `x in{9/4;3/4;25/4;-13/4}`
