$\frac{2b+2}{2b-b²}$÷$\frac{b+1}{b}$+$\frac{2a+2}{3b-6}$
=$\frac{2(b+1)}{b(2-b)}$×$\frac{b}{b+1}$+$\frac{2a+2}{3b-6}$
=$\frac{2}{2-b}$+$\frac{2a+2}{3(b-2)}$
=$\frac{2}{2-b}$-$\frac{2a+2}{3(2-b)}$ MTC:3(2-b)
=$\frac{2.3}{3(2-b)}$-$\frac{2a+2}{3(2-b)}$
=$\frac{6-(2a+2)}{3(2-b)}$
=$\frac{6-2a-2}{3(2-b)}$
=$\frac{4-2a}{3(2-b)}$