Question

tim x,y,z eZ (x-3)^2+/2y-6/+16z^2=0
e là thuộc

Answer 1

Ta có : (x - 3)² \(\ge0\forall x\in Z\)

            |2y - 6| \(\ge0\forall x\in Z\)

            16z² \(\ge0\forall x\in Z\)

Mà : (x - 3)² + |2y - 6| + 16z² = 0

Nên : \(\hept{\begin{cases}\left(x-3\right)^2=0\\\left|2y-6\right|=0\\16z^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-3=0\\2y-6=0\\z^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\2y=6\\z=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=3\\z=0\end{cases}}\)

Vậy x = 3 , y = 3 , z = 0 .