Question

[áddsa]

cho x,y,z khác nhau và khác 0 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

CM : \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}=0\)

Answer 1

[Nguyễn Thị Ngọc Thơ]

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\) \(\Rightarrow xy+yz+zx=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=-\left(yz+zx\right)\\yz=-\left(xy+zx\right)\\zx=-\left(xy+yz\right)\end{matrix}\right.\)

Thay vào ta có:

\(\frac{1}{x^2+2yz}=\frac{1}{x^2+yz+yz}=\frac{1}{x^2-xy+yz-zx}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)

CMTT:

\(PT\Leftrightarrow\frac{1}{\left(x-y\right)\left(x-z\right)}+\frac{1}{\left(x-y\right)\left(z-y\right)}+\frac{1}{\left(z-y\right)\left(z-x\right)}\)

\(\Leftrightarrow\frac{\left(z-y\right)+\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(z-y\right)}=0\left(đpcm\right)\)