2. Упростить выражение: a) $$\frac{3-2a}{2a}-\frac{1-a^2}{a^2}$$ б) $$\frac{1}{3x+y}-\frac{1}{3x-y}$$ в) $$\frac{4-3b}{b^2-2b} + \frac{3}{b-2}$$ г) $$\frac{2}{x-4}-\frac{x+8}{x^2-16} - \frac{1}{x}$$ д) $$\frac{3x+y}{y} \cdot (\frac{y}{x} - \frac{3y}{3x+y})$$ e) $$(\frac{st}{s^2-t^2} + \frac{t}{2t-2s}) \cdot \frac{s^2-t^2}{2t}$$

Решим каждое выражение по отдельности: а) $$ \frac{3-2a}{2a} - \frac{1-a^2}{a^2} = \frac{a(3-2a) - 2(1-a^2)}{2a^2} = \frac{3a - 2a^2 - 2 + 2a^2}{2a^2} = \frac{3a-2}{2a^2} $$ б) $$ \frac{1}{3x+y} - \frac{1}{3x-y} = \frac{(3x-y)-(3x+y)}{(3x+y)(3x-y)} = \frac{3x-y-3x-y}{(3x)^2 - y^2} = \frac{-2y}{9x^2 - y^2} $$ в) $$ \frac{4-3b}{b^2-2b} + \frac{3}{b-2} = \frac{4-3b}{b(b-2)} + \frac{3}{b-2} = \frac{4-3b + 3b}{b(b-2)} = \frac{4}{b(b-2)} = \frac{4}{b^2 - 2b} $$ г) $$ \frac{2}{x-4} - \frac{x+8}{x^2-16} - \frac{1}{x} = \frac{2}{x-4} - \frac{x+8}{(x-4)(x+4)} - \frac{1}{x} = \frac{2x(x+4) - x(x+8) - (x-4)(x+4)}{x(x-4)(x+4)} = \frac{2x^2 + 8x - x^2 - 8x - (x^2-16)}{x(x^2-16)} = \frac{2x^2 - x^2 - x^2 + 8x - 8x + 16}{x(x^2-16)} = \frac{16}{x(x^2-16)} $$ д) $$ \frac{3x+y}{y} \cdot (\frac{y}{x} - \frac{3y}{3x+y}) = \frac{3x+y}{y} \cdot \frac{y(3x+y) - 3xy}{x(3x+y)} = \frac{3x+y}{y} \cdot \frac{3xy + y^2 - 3xy}{x(3x+y)} = \frac{3x+y}{y} \cdot \frac{y^2}{x(3x+y)} = \frac{y}{x} $$ e) $$ (\frac{st}{s^2-t^2} + \frac{t}{2t-2s}) \cdot \frac{s^2-t^2}{2t} = (\frac{st}{(s-t)(s+t)} - \frac{t}{2(s-t)}) \cdot \frac{(s-t)(s+t)}{2t} = \frac{2st - t(s+t)}{2(s-t)(s+t)} \cdot \frac{(s-t)(s+t)}{2t} = \frac{2st - st - t^2}{2(s-t)(s+t)} \cdot \frac{(s-t)(s+t)}{2t} = \frac{st - t^2}{2(s-t)(s+t)} \cdot \frac{(s-t)(s+t)}{2t} = \frac{t(s-t)}{2(s-t)(s+t)} \cdot \frac{(s-t)(s+t)}{2t} = \frac{s-t}{4} $$