Let C be a complex of right A-modules and E a complex of left A-modules. We assume that H(C)=0 and that E is flat and bounded on the right. Then H(C⊗AE)=0.
For k∈Z, let T(k) be the subcomplex of C⊗AE such that
Tn(k)=p+q=nq≤k∑Cp⊗AEq;
then T(k−1)⊂T(k) and we have an exact sequence of complexes
0⟶T(k−1)ikT(k)πC⊗AEk(−k)⟶0
where ik is the canonical injection and where π projects the preceding direct sum onto its factor Cn−k⊗AEk=(C⊗AEk(−k))n. By Cor. 2 above, we have H(C⊗AEk(−k))=0, hence ik is a homomorphism. We have T(k)=0 for k sufficiently small, since E is bounded on the right, hence H(T(k))=0 for all k by induction on k. Finally, the canonical morphism ⟶limT(k)→C⊗AE is obviously an isomorphism, hence H(C⊗AE)=0 (X, p. 28, Prop. 1).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.