a) Ta có
$x^2 - 25 = (2x-1)(x+5)$
$<-> x^2 -25 = 2x^2 +9x - 5$
$<-> x^2 +9x +20 = 0$
$<-> (x+4)(x+5) = 0$
Vậy $x = -4$ hoặc $x = -5$
b) ĐK: $x \neq 0, -2$. Ta có
$\dfrac{x-2}{x+2} - \dfrac{x^2+2}{x^2 + 2x} = \dfrac{3}{x}$
$<-> \dfrac{x(x-2)}{x(x+2)} - \dfrac{x^2 + 2}{x(x+2)} = \dfrac{3(x+2)}{x(x+2)}$
$<-> x^2 - 2x - (x^2 +2) = 3x + 6$
$<-> 5x = -8$
$<-> x = -\dfrac{8}{5}$