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\(\frac{zx}{yz}=\frac{1}{2}\Rightarrow\frac{x}{y}=\frac{1}{2}\)
\(\frac{x}{yz}:\frac{y}{xz}=\frac{x}{yz}.\frac{xz}{y}=\frac{x^2}{y^2}\)
Mà \(\frac{x}{y}=\frac{1}{2}\Rightarrow\left(\frac{x}{y}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
Vậy \(\frac{x}{yz}:\frac{y}{zx}=\frac{1}{4}\)
\(\frac{yz}{xz}=\frac{1}{2}\Rightarrow\frac{y}{x}=\frac{1}{2}\)
\(\frac{x}{yz}:\frac{y}{xz}=\frac{x}{yz}\cdot\frac{xz}{y}==\frac{x^2}{y^2}\)
Vì \(\frac{y}{x}=\frac{1}{2}\Rightarrow\frac{x}{y}=2\Rightarrow\frac{x^2}{y^2}=2^2=4\)
=> x/yz : y/xz = 4
Ta có: \(\frac{yz}{zx}=\frac{1}{2}\Rightarrow2yz=zx\Rightarrow2y=x\Rightarrow\frac{x}{y}=2\)
\(\frac{x}{yz}:\frac{y}{zx}=\frac{x^2z}{y^2z}=\frac{x^2}{y^2}=\left(\frac{x}{y}\right)^2=2^2=4\)
Vậy \(\frac{x}{yz}:\frac{y}{zx}=4\)
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
ai tl giup t di ma