b) Ta có
$(2x+5)(x-4) - (x-5)(4-x) = 0$
$<-> (2x+5)(x-4) + (x-5)(x-4) = 0$
$<-> (x-4)(2x+5 + x-5) = 0$
$<-> (x-4)3x = 0$
Vậy $x = 0$ hoặc $x = 4$.
d) Ta có
$2(9x^2 + 6x + 1) = (3x+1)(x-2)$
$<-> 2(3x+1)^2 - (3x+1)(x-2) = 0$
$<-> (3x+1)[2(3x+1) - (x-2)] = 0$
$<-> (3x+1)(5x +4) = 0$
Vậy $x = -\dfrac{1}{3}$ hoặc $x = -\dfrac{4}{5}$.
f) Ta có
$16x^2 - 8x + 1 = 4(x+3)(4x-1)$
$<-> (4x-1)^2 - 4(x+3)(4x-1) = 0$
$<-> (4x-1)[(4x-1)-4(x+3)] = 0$
$<-> (4x-1)(-13) = 0$
Vậy $x = \dfrac{1}{4}$.