Có:
[tex]A=\frac{6xy}{z^2}+\frac{3yz}{4x^2}+\frac{2zx}{9y^2}=\frac{6xyz}{z^3}+\frac{3xyz}{4x^3}+\frac{2xyz}{9y^3}=\frac{6xyz}{z^3}+\frac{6xyz}{(2x)^3}+\frac{6xyz}{(3y)^3}=6xyz(\frac{1}{z^3}+\frac{1}{(2x)^3}+\frac{1}{(3y)^3})=6xyz((\frac{1}{z}+\frac{1}{2x}+\frac{1}{3y})(\frac{1}{z^2}+\frac{1}{4x^2}+\frac{1}{9y^2}-\frac{1}{2xz}-\frac{1}{3yz}-\frac{1}{6xyz})+\frac{3}{6xyz})=6xyz(0+\frac{3}{6xyz}))=3[/tex]