a) 4x(x + 1) = 8(x + 1)
=> 4x(x + 1) - 8(x + 1) = 0
=> 4(x + 1).(x - 2) = 0
=> (x + 1)(x - 2) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\x=2\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=-1\\x=2\end{array}\right.\)
b) x2(x - 2) + 2 - x = 0
=> x2.(x - 2) - (x - 2) = 0
=> (x - 2).(x2 - 1) = 0
=> (x - 2).(x - 1).(x + 1) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\\x+1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=1\\x=-1\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=2\\x=1\\x=-1\end{array}\right.\)