\(\sqrt{2}\left(x-z\right)+y-2=\sqrt{2}\left(y+1\right)-3x+z\)
\(\Leftrightarrow3x+y-z-2=\sqrt{2}\left(-x+y+z+1\right)\)
Vì \(x;y;z\in Z^+\) và \(\sqrt{2}\) là số vô tỷ
\(\Rightarrow\left\{{}\begin{matrix}-x+y+z+1=0\\3x+y-z-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=1\\-x+y+z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\z=x-y-1\end{matrix}\right.\)
\(P=\dfrac{x+y+z}{2y-1}=\dfrac{\dfrac{1}{2}+x-y-1}{2y-1}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2}-y-y-1}{2y-1}\) \(\left(x=\dfrac{1}{2}-y\right)\)
\(\Rightarrow P=\dfrac{-2y}{2y-1}=-\dfrac{2y}{2y-1}\)
mà \(2y-1;2y\) là 2 số nguyên dương liên tiếp nên \(\left(2y-1;2y\right)=1\)
\(\)\(\Rightarrow\dfrac{2y}{2y-1}\) hay \(P=-\dfrac{2y}{2y-1}\) là phân số tối giản
\(\Rightarrowđpcm\)
