Ans:

$-$ Cho $H(x) = 0$:

$2x^3 - 2x^2 - 6x = 0$

$2x(x^2 - x - 3) = 0$

$→$ TH1: $2x = 0 \implies x = 0$

$→$ TH2: $x^2 - x - 3 = 0 \implies x = \dfrac{1 \pm \sqrt{13}}{2}$

⇒ Nghệm: $x = 0$, $x = \dfrac{1 + \sqrt{13}}{2}$, $x = \dfrac{1 - \sqrt{13}}{2}$

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