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Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2019}{y}=\frac{x+y-2020}{z}=\frac{y+z+1+x+z+2019+x+y-2020}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow2=\frac{1}{x+y+z}\)\(\Rightarrow x+y+z=\frac{1}{2}\)
Ta có:
+) \(\frac{y+z+1}{x}=2\)\(\Rightarrow y+z+1=2x\)\(\Rightarrow x+y+z+1=3x\)\(\Rightarrow\frac{1}{2}+1=3x\)\(\Rightarrow3x=\frac{3}{2}\)\(\Rightarrow x=\frac{1}{2}\)
+) \(\frac{x+z+2019}{y}=2\)\(\Rightarrow x+z+2019=2y\)\(\Rightarrow x+y+z+2019=3y\)\(\Rightarrow\frac{1}{2}+2019=3y\)\(\Rightarrow3y=\frac{4039}{2}\)\(\Rightarrow y=\frac{4039}{6}\)
+) \(\frac{x+y-2020}{z}=2\)\(\Rightarrow x+y-2020=2z\)\(\Rightarrow x+y+z-2020=3z\)\(\Rightarrow\frac{1}{2}-2020=3z\)\(\Rightarrow3z=\frac{-4039}{2}\)\(\Rightarrow z=\frac{-4039}{6}\)
Lại có: \(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{4039}{6}\right)^{2017}+\left(\frac{-4039}{6}\right)^{2017}=4032+\left(\frac{4039}{6}\right)^{2017}-\left(\frac{4039}{6}\right)^{2017}=4032\)
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