A,B decidable: proof that A\B is decidable too
Let $M_A$ and $M_B$ be deciders for $A$ and $B$ respectively. You can construct a TM $M$,

$M$ = "On input x,

$\qquad$ 1. Simulate $M_A$ on x.

$\qquad$ 2. Reject if $M_A$ rejects. Else simulate $M_B$ on x.

$\qquad$ 3. Accept if $M_B$ rejects. Else reject."

$Correctness$: 

$\qquad$ $M$ always halts as $M_A$ and $M_B$ always halts. $M$ accepts if $M_A$ accepts and $M_B$ rejects i.e, $x \in A$ and $x \notin B$. $M$ rejects otherwise.