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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
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)
Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)
\(\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{x+z}{y}=\dfrac{x^2y+xy^2+y^2z+yz^2+x^2z+xz^2}{xyz}=\dfrac{-3xyz}{xyz}=-3\)
đề cho xy+yz+xz=0 nhân cả 2 vế với -z
=>-xyz-\(z^2\left(y+x\right)\)=0
=>-xyz=\(z^2x+z^2y\)
cmtt bạn nhân với -y và -z
=>-3xyz=\(x^2y+xy^2+y^2z+yz^2+x^2z+xz^2\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow xy+yz+xz=0\)
\(A=\frac{yz}{x^2+yz+-xy-xz}+\frac{xz}{y^2+zx-xy-yz}+\frac{xy}{z^2+xy-xz-yz}\)
\(A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-z\right)\left(y-x\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(A=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)
\(A=\frac{\left(z-x\right)\left(y-z\right)\left(y-x\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}=1\)