Đáp án:
Ta có: `f (x) = g (x)`
`⇔ 4x^3 - 2x^2 + x - 5 = 4x^3 + 4x^2 - 5x - 5`
`⇔ 4x^3 - 2x^2 + x - 5 - 4x^3 - 4x^2 + 5x + 5 = 0`
`⇔ (4x^3 - 4x^3) + (-2x^2 - 4x^2) + (x + 5x) + (-5 + 5) = 0`
`⇔ - 6x^2 + 6x = 0`
`⇔ 6x (- x + 1 ) = 0`
`⇒` $\left[\begin{matrix} 6x=0\\ -x + 1=0\end{matrix}\right.$
`⇒` $\left[\begin{matrix} x = 0\\ - x= - 1\end{matrix}\right.$
`⇒` $\left[\begin{matrix} x=0\\ x = 1\end{matrix}\right.$
Vậy `x = 0` hoặc `x = 1`