Để `(2x+3)/7 in Z` `=>2x+3 vdots 7` `=>2x+3 in B(7)` `B(7)={...-14; -7; 0; 7; 14; ...}` `=>2x+3 in {...-14; -7; 0; 7; 14; ...}` `=>2x in {...-17; -10; -3; 4; 11; ...}` `=>x in {...(-17)/2; -5; (-3)/2; 2; (11)/2; ...}` Mà `x in Z` `=>x in {...-5; 2; ...}` Vậy : `x in {...-5; 2; ...}` thì `(2x+3)/7 in Z`
Để `(2x+3)/7\inZZ` `=>2x+3\vdots7` `=>2x+3\in B(7)` Mà `B(7)={0;7;-7;14;-14;21;-21;....}` `=>2x\in{-3;4;-10;11;-17;18;-24;...}` `=>x\in{-3/2;2;-5;11/2;-17/2;9;-12;...}` Vậy `x\in{-3/2;2;-5;11/2;-17/2;9;-12;...}` thì `(2x+3)/7` là số nguyên
