\(< =>\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(< =>\left\{{}\begin{matrix}x-2000=1\\y-2001=1\\z-2002=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2001\left(TM\right)\\x=2002\left(TM\right)\\x=2002\left(TM\right)\end{matrix}\right.\)
\(2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=x+y+z-6000\)
\(\Leftrightarrow x-2000-2\sqrt{x-2000}+1+y-2001-2\sqrt{y-2001}+1+z-2002-2\sqrt{z-2002}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x-2000=1\\y-2001=1\\z-2002=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
Điều kiện xác định: \(\left\{{}\begin{matrix}x-2000\ge0\\y-2001\ge0\\z-2002\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2000\\y\ge2001\\z\ge2002\end{matrix}\right.\)
\(\sqrt{x-2000}+\sqrt{y-2001}+\sqrt{z-2002}=\dfrac{1}{2}\left(x+y+z\right)-3000\)
\(\Leftrightarrow2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=\left(x+y+z\right)-6000\)
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2000}-1\right)^2=0\\\left(\sqrt{y-2001}-1\right)^2=0\\\left(\sqrt{z-2002}-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}=1\\\sqrt{y-2001}=1\\\sqrt{z-2002}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2000=1\\y-2001=1\\z-2002=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\) (nhận)
Vậy \(\left(x;y;z\right)=\left(2001;2002;2003\right)\) là nghiệm của phương trình.
