`6x^4 – 5x^3 – 38x^2 – 5x + 6 = 0`
`⇔ 6x^4 – 18x^3 + 13x^3 – 39x^2 + x^2 – 3x - 2x + 6 = 0`
`⇔ 6x^3 (x - 3) + 13x^2 (x - 3) + x (x - 3) - 2 (x - 3) = 0`
`⇔ (6x^3 + 13x^2 + x - 2) (x - 3) = 0`
`⇔ (6x^3 + 12x^2 + x^2 + 2x - x - 2) (x - 3) = 0`
`⇔ [6x^2 (x + 2) + x (x + 2) - (x + 2)] (x - 3) = 0`
`⇔ (6x^2 + x - 1) (x + 2) (x - 3) = 0`
`⇔ (6x^2 - 2x + 3x - 1) (x + 2) (x - 3) = 0`
`⇔ [2x (3x - 1) + (3x - 1)] (x + 2) (x - 3) = 0`
`⇔ (2x + 1) (3x - 1) (x + 2) (x - 3) = 0`
`⇔` \(\left[ \begin{array}{l}2x + 1=0\\3x - 1 = 0\\ x + 2 = 0\\x - 3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x = - 1\\3x = 1\\ x + 2 = 0\\x - 3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = \dfrac{-1}{2} \\x = \dfrac{1}{3}\\ x = - 2\\x =3\end{array} \right.\)
Vậy `x ∈ ` {$\dfrac{-1}{2}$ ; $\dfrac{1}{3}$ ,` - 2; 3}`
