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Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)
\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)
\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)
\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)
Vậy A = 2019
\(P=\frac{2019xz}{xyz+2019xz+2019z}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{2019xz}{2019+2019xz+2019z}+\frac{y}{y\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)
\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=1\)
xin loi , may tinh minh hong unikey
Dat \(\frac{x}{2017}=\frac{y}{2018}=\frac{z}{2019}=k\)
Suy ra \(x=2017k;y=2018k;z=2019k\)
Khi đó 4.(x-y).(y-z) = \(4.\left(2017k-2018k\right).\left(2018k-2019k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)
\(\left(z-x\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)
Nen \(4.\left(x-y\right).\left(y-z\right)=\left(z-x\right)^2\)