Question

Помогите, не получается
10 баллов даю

Answer 1

Здравствуйте!

Решение вашей задачи ниже:

$\frac{2-5p-2k+5pk}{10p^{2}-9+2 } = \frac{2-5p2k+5kp}{10p^{2}-4p-5p+2 } = \frac{2p-5p-k(2-5p)}{2p(5p-2)-(5p-2)} = \frac{(2-5p)(1-k)}{(5p-2)(2p-1)} = \frac{-(5p-2)(1-k)}{(5p-2)(2p-1)} =\\\frac{-(1-k)}{(2p-1)} = -\frac{1-k}{2p-1} = \frac{k-1}{2p-1}$

Answer 2

$\tt \cfrac{2-5p-2k+5pk}{10p^2-9p+2}=\cfrac{2-5p-k(2-5p)}{10p^2-4p-5p+2}=\cfrac{(2-5p)(1-k)}{2p(5p-2)-(5p-2)}=\\\\\\ =\cfrac{-(5p-2)(1-k)}{(5p-2)(2p-1)}=\cfrac{-(1-k)}{2p-1}=\cfrac{k-1}{2p-1}$