Упростите выражение:

• 1) $$\frac{y+3}{2y+2} - \frac{y+1}{2y-2} + \frac{3}{y^2-1} = \frac{y+3}{2(y+1)} - \frac{y+1}{2(y-1)} + \frac{3}{(y-1)(y+1)} = \frac{(y+3)(y-1) - (y+1)(y+1) + 6}{2(y-1)(y+1)} = \frac{y^2+2y-3 - (y^2+2y+1) + 6}{2(y-1)(y+1)} = \frac{y^2+2y-3 - y^2 - 2y - 1 + 6}{2(y-1)(y+1)} = \frac{2}{2(y-1)(y+1)} = \frac{1}{(y-1)(y+1)} = \frac{1}{y^2-1}$$
• 2) $$\frac{2b^2-b}{b^3+1} - \frac{b-1}{b^2-b+1} = \frac{2b^2-b}{(b+1)(b^2-b+1)} - \frac{b-1}{b^2-b+1} = \frac{2b^2-b - (b-1)(b+1)}{(b+1)(b^2-b+1)} = \frac{2b^2-b - (b^2-1)}{(b+1)(b^2-b+1)} = \frac{2b^2-b - b^2 + 1}{(b+1)(b^2-b+1)} = \frac{b^2 - b + 1}{(b+1)(b^2-b+1)} = \frac{1}{b+1}$$