Vì \(\left(\sqrt{x};x-\sqrt{x}+1\right)=1\) nên \(M\in Z\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-\sqrt{x}+1}\in Z\\x\in N\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\sqrt{x}+1=-2\\x-\sqrt{x}+1=-1\\x-\sqrt{x}+1=1\\x-\sqrt{x}+1=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-\sqrt{x}+3=0\left(VN\right)\\x-\sqrt{x}+2=0\left(VN\right)\\x-\sqrt{x}=0\left(1\right)\\x-\sqrt{x}-1=0\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
(2) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{1+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{1-\sqrt{5}}{2}\left(vô-lý\right)\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{3+\sqrt{5}}{2}\left(L\right)\)
Kl: x=0, x=1
