Question

a,x+6 chia hết cho x
b,x+9 chia hết cho x+1
c,2x+1 chia hết cho x-1

Answer 1

a: \(\Leftrightarrow6⋮x\)

hay \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

b: =>x+1+8 chia hết cho x+1

\(\Leftrightarrow x+1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

hay \(x\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)

c: =>2x-2+3 chia hết cho x-1

\(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)