Доказать, что при x+y+z=0 [latex](\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z})*(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y})=9[/latex],    

[latex]x+y+z=0\\ y=-(x+z)\\ (\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z})(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y})=9\\ (\frac{-(x+z)-z}{x}+\frac{z-x}{-(x+z)}+\frac{2x+z}{z})(\frac{x}{-(x+z)-z}+\frac{-(x+z)}{z-x}+\frac{z}{2x+z})=9\\ (\frac{-x-2z}{x}+\frac{z-x}{-x-z}+\frac{2x+z}{z})(\frac{x}{-x-2z}+\frac{-x-z}{z-x}+\frac{z}{2x+z})=9\\ zamena \\ \frac{-x-2z}{x}=a;\\ \frac{z-x}{-x-z}=b;\\ \frac{2x+z}{z}=c;\\ togda\\ \frac{x}{-x-2z}=\frac{1}{a}\\ \frac{-x-z}{z-x}=\frac{1}{b}\\ \frac{z}{2x+z}=\frac{1}{c}\\ [/latex] Затем  [latex](a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{(a+b+c)(ab+bc+ac)}{abc};\\ podstavim \ nashi\ zna4enia\ \\ poluchim \\ 9(-\frac{2x}{z}}-\frac{-2x}{x+z}+\frac{2x}{z}+1) \\ v\ znamenatele \ polu4im \ [/latex] в знаменателе  [latex](\frac{-2x}{z}-\frac{2x}{x+z}+\frac{2z}{x}+1)[/latex] в итоге получим 9