Ta có: \(\left(5x-7\right)^2\ge0\) \(\left(6-5y\right)^2\ge0\) \(\left(5z\right)^2\ge0\) Nên để \(\left(5x-7\right)^2+\left(6-5y\right)^2+\left(5z\right)^2=0\)
Mình bổ sung: \(\Rightarrow\left(5x-7\right)^2=0\) ; \(\left(6+5y\right)^2=0\) \(\left(5z\right)^2=0\) \(\rightarrow\) 5x-7=0 . \(\rightarrow5x=7\Rightarrow x=\dfrac{7}{5}\) \(\left(6-5y\right)^2=0\rightarrow6-5y=0\rightarrow5y=6\rightarrow y=\dfrac{6}{5}\)
\(\left(5z\right)^2=0\rightarrow5z=0\) \(\rightarrow z=0\) Vậy x=\(\dfrac{7}{5}\) ; z=0; y= \(\dfrac{6}{5}\)
Ta có:\(\left\{{}\begin{matrix}\left(5x-7\right)^2\ge0\\\left(6-5y\right)^2\ge0\\\left(5z\right)^2\ge0\end{matrix}\right.\) với mọi x, y, z \(\Rightarrow\left(5x-7\right)^2+\left(6-5y\right)^2+\left(5z\right)^2\ge0\) \(\Rightarrow\left(5x-7\right)^2+\left(6-5y\right)^2+\left(5z\right)^2=0\) khi
\(\left\{{}\begin{matrix}\left(5x-7\right)^2=0\\\left(6-5y\right)^2=0\\\left(5z\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}5x-7=0\\6-5y=0\\5z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}\\y=\dfrac{6}{5}\\z=0\end{matrix}\right.\) (ko thỏa mãn)
Vậy không tồn tại các số nguyên x, y thỏa mãn yêu cầu bài toán.
Chúc p hk tốt 
