Đặt \(A=x^4+\left(x+1\right)^3-2x^2-2x=x^4+x^3+x^2+x+1\)
\(4A=4x^4+4x^3+4x^2+4x+4=\left(2x^2+x\right)^2+2x^2+\left(x+2\right)^2>\left(2x^2+x\right)^2\)
\(4A=\left(2x^2+x\right)^2+4\left(2x^2+x\right)+4-5x^2=\left(2x^2+x+2\right)^2-5x^2\le\left(2x^2+x+2\right)^2\)
\(\Rightarrow\left(2x^2+x\right)^2< 4A\le\left(2x^2+x+2\right)^2\)
Mà 4A là số chính phương \(\Rightarrow\left[{}\begin{matrix}4A=\left(2x^2+x+1\right)^2\\4A=\left(2x^2+x+2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\5x^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=0\end{matrix}\right.\)