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\(\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{\frac{1}{x+y+x}}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
B=\(\left(x+y\right)\left(y+z\right)\left(z+x\right).M=0\)
Ta có: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)
=>\(\frac{yz + zx + xy}{xyz}=0\)
=>xy+yz+xz=0
=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz
\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)
\(=x^2-xy-xz+yz\)
\(=x(x-y)-z(x-y)=(x-y)(x-z)\)
Chứng minh tương tự, ta sẽ có:
\(y^2 + 2zx = (y - z)(y - x)\)
\(z^2 + 2xy = (z - x)(z - y)\)
Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)
\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)
\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)
\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)
=0
=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)
\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)
\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)
=0
=xy+yz+xz