a, Ta có: 10-x$\vdots$x-3
⇒-(x-3)+7$\vdots$x-3
⇒x-3∈Ư(7)={±1;±7}
x-3=1⇒x=4
x-3=-1⇒x=2
x-3=7⇒x=10
x-3=-7⇒x=-4
Vậy x∈{4;2;10;-4}
b, Ta có: 2x+5$\vdots$x-1
⇒2(x-1)+7$\vdots$x-1
⇒x-1∈Ư(7)={±1;±7}
x-1=1⇒x=2
x-1=-1⇒x=0
x-1=7⇒x=8
x-1=-7⇒x=-6
Vậy x∈{2;0;8;-6}