Амалдарды орындандар: a) (m - 2n) : (2n - m) / 5mn; б) (a+5b)/(a2-5ab) : (25b2-a2)/5b2

The denominators are $4x-y$ and $y-4x$. Notice that $y-4x = -(4x-y)$.

$$\frac{3x+2}{4x-y} + \frac{2x-1}{y-4x} = \frac{3x+2}{4x-y} + \frac{2x-1}{-(4x-y)}$$

$$\frac{x+3}{4x-y}$$

The final answer is $\boxed{\frac{x+3}{4x-y}}$.

б)

$$a^2-5ab = a(a-5b)$$ $$a^2+5ab = a(a+5b)$$

The expression becomes:

$$\left( \frac{a+5b}{a(a-5b)} - \frac{a-5b}{a(a+5b)} \right) : \frac{25b^2-a^2}{5b^2}$$

$$\left( \frac{(a+5b)(a+5b)}{a(a-5b)(a+5b)} - \frac{(a-5b)(a-5b)}{a(a-5b)(a+5b)} \right) : \frac{25b^2-a^2}{5b^2}$$ $$\left( \frac{(a+5b)^2 - (a-5b)^2}{a(a-5b)(a+5b)} \right) : \frac{25b^2-a^2}{5b^2}$$

$$(a+5b)^2 - (a-5b)^2 = ((a+5b)-(a-5b))((a+5b)+(a-5b))$$ $$= (a+5b-a+5b)(a+5b+a-5b) = (10b)(2a) = 20ab$$

The expression inside the parenthesis simplifies to:

$$\frac{20ab}{a(a-5b)(a+5b)} = \frac{20b}{(a-5b)(a+5b)} = \frac{20b}{a^2-25b^2}$$

Division by a fraction is multiplication by its reciprocal. Also, note that $25b^2-a^2 = -(a^2-25b^2)$.

$$\frac{20b}{a^2-25b^2} : \frac{25b^2-a^2}{5b^2} = \frac{20b}{a^2-25b^2} \times \frac{5b^2}{25b^2-a^2}$$ $$= \frac{20b}{a^2-25b^2} \times \frac{5b^2}{-(a^2-25b^2)}$$

$$\frac{20b \cdot 5b^2}{-(a^2-25b^2)(a^2-25b^2)} = \frac{100b^3}{-(a^2-25b^2)^2}$$ $$= -\frac{100b^3}{(a^2-25b^2)^2}$$

The final answer is $\boxed{-\frac{100b^3}{(a^2-25b^2)^2}}$.