\(\dfrac{x}{y}=\dfrac{2017}{2}\left(1\right)\\ \dfrac{y}{z}=\dfrac{2}{2017}\left(2\right)\\ x-2z=2017\left(3\right)\)
ĐK: \(y,z\ne0\)
Từ (1),(2) \(\Rightarrow\dfrac{x}{y}.\dfrac{y}{z}=\dfrac{2017}{2}.\dfrac{2}{2017}=1\Rightarrow\dfrac{x}{z}=1\Rightarrow x=z\)
Thay vào (3) \(\Rightarrow z-2z=2017\Rightarrow z=-2017\)
Từ (1) \(\Rightarrow y=-2\)
KL:
\(x=-2017\\
y=-2\\
z=-2017\)
Ta có \(\dfrac{x}{y}=\dfrac{2017}{2};\dfrac{y}{z}=\dfrac{2}{2017}\)
=>2x=2017y;2z=2017y
=>2z=2x
=>x=z
=>x-2z=z-2z=-z
mà x-2z=2017
=>-z=2017
=>z=-2017
=>x=-2017
=>2017y=(-2017).2
=>y=-2.2017:2017
=>y=-2
Từ \(\dfrac{x}{y}=\dfrac{2017}{2};\dfrac{y}{z}=\dfrac{2}{2017}\)
=>\(\dfrac{x}{2017}=\dfrac{y}{2}=\dfrac{z}{2017}\)
Áp dụng t/c của dãy tỉ sô bằng nhau,ta có:
\(\dfrac{x}{2017}=\dfrac{y}{2}=\dfrac{z}{2017}=\dfrac{x-2z}{2017-2.2017}=\dfrac{2017}{-2017}=-1\)
=>\(x=-1.2107=-2017\)
\(y=-1.2=-2\)
\(z=-1.2017=-2017\)
