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H

hiendang241

ádfghjk

ta có $\frac{1}{x}$+$\frac{1}{y}$+$\frac{1}{z}$=0
nên xy+yz+zx=0\Rightarrow xy=-xz-yz;xz=-xy-yz;yz=-xy-yz
\Rightarrow $\frac{xy}{z^2+2xy}$ +$\frac{zy}{x^2+2yz}$+$\frac{xz}{y^2+2xz}$
=$\frac{xy}{z^2+xy-xz-yz}$+$\frac{yz}{x^2+yz-xy-xz}$+$\frac{xz}{y^2+xz-xy-yz}$
=$\frac{xy}{(y-z)(x-z)}$+$\frac{yz}{(x-y)(x-z)}$+$\frac{xz}{(z-y)(x-y)}$
=$\frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}$
=1
 
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