Mình đọc chưa hết đề nên làm thiếu, cậu bổ sung nhé:
Thay vào P, ta có:
\(P=2017x+y^{2018}+z^{2019}=2017.\frac{1}{2}+\left(\frac{5}{6}\right)^{2018}+\left(\frac{-5}{6}\right)^{2019}=1008,5+\frac{5^{2018}}{6^{2018}}.\frac{1}{6}=1008,5+\frac{5^{2018}}{6^{2019}}\)
điều kiện x,y,z khác 0
ta có \(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}\\ =\frac{y+z+1+z+x+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\\ \Rightarrow x+y+z=\frac{1}{2}\\ \Rightarrow y+z=\frac{1}{2}-x\left(1\right)\)
\(\frac{y+z+1}{x}=2\\ \Leftrightarrow y+z+1=2x\)
kết hợp với (1)
có \(\frac{1}{2}-x+1=2x\\ \Leftrightarrow2x+x=\frac{1}{2}+1\\ \Leftrightarrow3x=\frac{3}{2}\\ \Leftrightarrow x=\frac{1}{2}\)
mà x + y + z = \(\frac{1}{2} \)
=> y + z = 0
=> y = -z
có \(\frac{x+y-3}{z}=2\\ \Leftrightarrow x+y-3=2z\\ \Leftrightarrow y-z=-\frac{5}{2}\)
mà y = -z
=> \(-3z=-\frac{5}{2}\\ \Rightarrow z=\frac{5}{6}\)
=> y = \(-\frac{5}{6}\)
=> \(P=2017.\frac{1}{2}+\left(-\frac{5}{6}\right)^{2018}+\left(\frac{5}{6}\right)^{2019}\)
\(=1008,5+\left(\frac{5}{6}\right)^{2018}+\left(\frac{5}{6}\right)^{2019}\)
Ta có:
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+z+x+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2=\frac{1}{x+y+z}\)\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}y+z=\frac{1}{2}-x\\z+x=\frac{1}{2}-y\\x+y=\frac{1}{2}-z\end{matrix}\right.\)
\(\Rightarrow\frac{\frac{1}{2}-x+1}{x}=\frac{\frac{1}{2}-y+2}{y}=\frac{\frac{1}{2}-z-3}{z}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{matrix}\right.\)
