\(...\Leftrightarrow y=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(\Leftrightarrow y=\sqrt{\left(\sqrt{x-1}+1\right)^2+}\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(\Leftrightarrow y=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
Điều kiện \(\sqrt{x-1}\ge0\) \(\left(x\ge1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}+1\ge1\\\sqrt{x-1}-1\ge-1\end{matrix}\right.\)
\(\Rightarrow y=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\ge2\)
Dấu "=" xảy ra khi \(x=1\Rightarrow y=2\)
mà \(x;y\in Z;x\ge1\)
Nên nghiệm của phương trình cho là
\(\left\{{}\begin{matrix}y=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\\x\ge1;y\ge2\\x;y\in Z^+\end{matrix}\right.\)
