Cho [tex]x,y,z[/tex] đôi 1 khác nhau và khác 0 thỏa mãn: [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0[/tex]. Tính giá trị của [tex]A=\frac{yz}{x^2+2yz}+\frac{zx}{y^2+2xz}+\frac{xy}{z^2+2xy}[/tex]
Thanks
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} =0$
=> $xy+yz+zx=0$
=>$A=\frac{yz}{x^2+2yz}+\frac{zx}{y^2+2xz}+\frac{xy}{z^2+2xy} = \frac{yz}{x^2-xy-zx+zy+(xy+yz+zx)}+\frac{zx}{y^2+zx-xy-yz+(xy+yz+zx)}+\frac{xy}{z^2+xy-yz-zx+(xy+yz+zx)} = \frac{yz}{(x-y)(x-z)}+\frac{zx}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$ (Do xy+yz+zx=0)
=> $A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-zx(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{(x-y)(y-z)(z-x)} = \frac{z^{2}y-y^{2}z-z^{2}x+zx^{2}-x^{2}y+xy^{2}}{(x-y)(y-z)(z-x)} = \frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1$