Lời giải:
\(\frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}\)
\(\Rightarrow (\frac{x+y}{y+z})^4=(\frac{y+z}{z+t})^4=(\frac{z+t}{t+x})^4=(\frac{t+x}{x+y})^4=\frac{x+y}{y+z}.\frac{y+z}{z+t}.\frac{z+t}{t+x}.\frac{t+x}{x+y}=1\)
\(\Rightarrow \left[\begin{matrix} \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=1\\ \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=-1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=y=z=t\\ x+y+z+t=0\end{matrix}\right.\)
Nếu $x=y=z=t$ thì:
\(A=\left(\frac{y+z}{x+t}\right)^{2013}+\left(\frac{y+t}{x+y}\right)^{2014}=\left(\frac{x+x}{x+x}\right)^{2013}+\left(\frac{x+x}{x+x}\right)^{2014}=1+1=2\in\mathbb{Z}\)
Nếu $x+y+z+t=0$ thì:
\(y+z=-(x+t); y+t=-(x+y)\)
\(\Rightarrow A=(-1)^{2013}+(-1)^{2014}=(-1)+1=0\in\mathbb{Z}\)
Vậy biểu thức $A$ luôn có giá trị nguyên.
