1/(2х² + 5х) - 2/(25 - 10х) -4/(4х² - 25) = 1/5
1/[x(2x + 5)] - 2/[5(5 - 2x)] - 4/[(2x + 5)(2x - 5)] = 1/5
2x + 5 ≠ 0 ⇒ x ≠ -2,5
2x - 5 ≠ 0 ⇒ x ≠ 2,5
х ≠ 0
Cложим дроби: 1/[x(2x + 5)] - 4/[(2x + 5)(2x - 5) =
= (2х - 5 - 4х)/[x(2x + 5)(2x - 5)] =
= - (2x + 5)/[x(2x + 5)(2x - 5)] =
= 1/[x(5 - 2x)]
и ещё:
1/[x(5 - 2x)]- 2/[5(5 - 2x)] = (5 - 2x)/[5x(5 - 2x)] = 1/(5x)
вернёмся к уравнению
1/(5x) = 1/5
5x = 5
x = 1