m, Ta có: x-3$\vdots$2-x
⇒x-3$\vdots$-(x-2)
⇒(x-2)-1$\vdots$-(x-2)
⇒2-x∈Ư(1)={±1}
2-x=1⇒x=1
2-x=-1⇒x=3
Vậy x∈{1;3}
n, Ta có: 3x-2$\vdots$2x-1
⇒2(3x-2)$\vdots$2x-1
⇒6x-4$\vdots$2x-1
⇒3(2x-1)-1$\vdots$2x-1
⇒2x-1∈Ư(1)={±1}
2x-1=1⇒x=1
2x-1=-1⇒x=0
Vậy x∈{1;0}