Ta có
$x^2 + x + \dfrac{1}{4} = \left( x + \dfrac{1}{2} \right)^2$
Thay vào ta có
$\sqrt{x^2 - \dfrac{1}{4} + x + \dfrac{1}{2}} = \dfrac{1}{2} (2x^3 + x^2 + 2x + 1)$
$<-> \sqrt{x^2 + x + \dfrac{1}{4}} = \dfrac{1}{2} [x^2(2x + 1) + (2x+1)]$
$<-> x + \dfrac{1}{2} = \dfrac{1}{2} (2x+1)(x^2 + 1)$
$<-> 2x + 1 = (2x+1)(x^2 + 1)$
$<-> (2x+1)(x^2 + 1 - 1) = 0$
$<-> (2x+1)x^2 = 0$
Vậy $x = 0$ hoặc $x = -\dfrac{1}{2}$.
