Question

Tìm x, y, z biết rằng \(\dfrac{x}{y+z+2016}=\dfrac{y}{x+z+2017}=\dfrac{z}{x+y-4033}=x+y+z\)

Answer 1

\(\dfrac{x}{y+z+2016}=\dfrac{y}{x+z+2017}=\dfrac{z}{x+y-4033}\\ =\dfrac{x+y+z}{x+y+z+2016+2017-4033}=1\\ \Rightarrow x+y+z=1\\ \Rightarrow\left\{{}\begin{matrix}x+y=1-z\\x+z=1-y\\y+z=1-x\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{1-x+2016}=\dfrac{y}{1-y+2017}=\dfrac{z}{1-z-4033}=1\\ \Rightarrow\left\{{}\begin{matrix}x=2017-x\\y=2018-y\\z=-4032-z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2017}{2}\\y=1009\\z=-2016\end{matrix}\right.\)