\(5x=4y=2z\)
\(\Leftrightarrow\frac{5x}{20}=\frac{4y}{20}=\frac{2z}{20}\)
\(\Leftrightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{10}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}=\frac{y}{5}=\frac{z}{10}=\frac{x-y+z}{4-5+10}=\frac{-18}{8}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{4}=-2\\\frac{y}{5}=-2\\\frac{z}{10}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=-10\\z=-20\end{matrix}\right.\)
Lại có :
\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}\)
\(=\left(\frac{2}{-8}+\frac{5}{-10}+\frac{5}{-20}\right)^{2016}\)
\(=1\)
Vậy....