2) a) (x+y)^2/6y + (x-y)^2/12y - (x^2-y^2)/4y;

2) a) $$\frac{(x+y)^2}{6y} + \frac{(x-y)^2}{12y} - \frac{x^2-y^2}{4y} = \frac{2(x+y)^2 + (x-y)^2 - 3(x^2-y^2)}{12y} = \frac{2(x^2+2xy+y^2) + (x^2-2xy+y^2) - 3(x^2-y^2)}{12y} = \frac{2x^2+4xy+2y^2+x^2-2xy+y^2-3x^2+3y^2}{12y} = \frac{6y^2+2xy}{12y} = \frac{2y(3y+x)}{12y} = \frac{3y+x}{6}$$

Ответ: $$\frac{3y+x}{6}$$