Đáp án:
`a, (x+1)^2 = x+1`
`-> (x+1)^2 - (x+1) = 0`
`-> (x+1) . [(x+1) - 1] = 0`
`->` $\left[\begin{matrix} x+1=0\\ (x+1) - 1 = 0\end{matrix}\right.$
`->` $\left[\begin{matrix} x=-1\\ x+1 = 1\end{matrix}\right.$
`->` $\left[\begin{matrix} x=-1\\ x=0\end{matrix}\right.$
Vậy $\left[\begin{matrix} x=-1\\ x=0\end{matrix}\right.$
`b, (x^3 - 4x^2)-(x-4) = 0`
`-> x^3 - 4x^2 - (x - 4) = 0`
`-> x^2.(x-4) - (x-4) = 0`
`-> (x-4).(x^2-1) = 0`
`->` $\left[\begin{matrix} x-4=0\\ x^2 - 1=0\end{matrix}\right.$
`->` $\left[\begin{matrix} x=4\\ x^2 = 1\end{matrix}\right.$
`->` $\left[\begin{matrix} x=4\\ x=±1\end{matrix}\right.$
Vậy $\left[\begin{matrix} x=4\\ x=±1\end{matrix}\right.$
