(2x2 + x)2 - 4(2x2 + x) + 3 = 0
Đặt 2x2 + x = t, ta có:
t2 - 4t + 3 = 0
t2 - t - 3t + 3 = 0
t(t - 1) - 3(t - 1) = 0
(t - 1)(t - 3) = 0
\(\left[\begin{matrix}t-1=0\\t-3=0\end{matrix}\right.\)
\(\left[\begin{matrix}2x^2+x-1=0\\2x^2+x-3=0\end{matrix}\right.\)
\(\left[\begin{matrix}2x^2+2x-x-1=0\\2x^2-2x+3x-3=0\end{matrix}\right.\)
\(\left[\begin{matrix}2x\left(x+1\right)-\left(x+1\right)=0\\2x\left(x-1\right)+3\left(x-1\right)=0\end{matrix}\right.\)
\(\left[\begin{matrix}\left(x+1\right)\left(2x-1\right)=0\\\left(x-1\right)\left(2x+3\right)=0\end{matrix}\right.\)
\(\left[\begin{matrix}x+1=0\\2x-1=0\\x-1=0\\2x+3=0\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\2x=1\\x=1\\2x=-3\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\x=\frac{1}{2}\\x=1\\x=-\frac{3}{2}\end{matrix}\right.\)
mà \(x\in Z\)
=> x = 1
