Question

Answer 1

1:

a: =>6a-2+3 chia hết cho 3a-1

=>\(3a-1\in\left\{1;-1;3;-3\right\}\)

mà a nguyên

nên a=0

b: \(\Leftrightarrow6a+10⋮2a-1\)

=>6a-3+13 chia hết cho 2a-1

=>\(2a-1\in\left\{1;-1;13;-13\right\}\)

=>\(a\in\left\{1;0;7;-6\right\}\)

c: \(\Leftrightarrow a^2-2a-3a+6-6⋮a-2\)

=>\(a-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(a\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

d: \(\Leftrightarrow6a-3-1⋮2a-1\)

=>\(2a-1\in\left\{1;-1\right\}\)

=>\(a\in\left\{1;0\right\}\)

e: \(\Leftrightarrow6a-9⋮3a+1\)

=>\(6a+2-11⋮3a+1\)

=>\(3a+1\in\left\{1;-1;11;-11\right\}\)

=>\(a\in\left\{0;-4\right\}\)