Ptrinh tương đương vs
$\sqrt{x^2 - \dfrac{1}{4} + \sqrt{x^2 + x + \dfrac{1}{4}}} = \dfrac{1}{2} (2x^3 + x^2 + 2x + 1)$
$<-> \sqrt{x^2 - \dfrac{1}{4} + \sqrt{(x + \dfrac{1}{2})^2}} = \dfrac{1}{2} (2x+1)(x^2 +1)$
$<-> \sqrt{x^2 + x + \dfrac{1}{4}} = \dfrac{1}{2} (2x+1)(x^2 + 1)$
$<-> 2\sqrt{( x + \dfrac{1}{2})^2} = (2x+1)(x^2+1)$
$<-> 2x + 1 = (2x+1)(x^2 + 1)$
$<-> (2x+1)(x^2 + 1 - 1) = 0$
Vậy $x = -\dfrac{1}{2}$ hoặc $x = 0$