a+b+c=0 -> a+b=-c và b+c=-a và c+a=-b
Ta có: [tex](\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a})(\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a})=(\frac{a-b}{-c}+\frac{b-c}{-a}+\frac{c-a}{-b})(\frac{-c}{a-b}+\frac{-a}{b-c}+\frac{-b}{c-a})=-(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}).(-(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}))=(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}). (\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a})[/tex]
Đó giờ đặt [tex]\frac{a-b}{c}=x \rightarrow \frac{c}{a-b}=\frac{1}{x}[/tex]
[tex]\frac{b-c}{a}=y \rightarrow \frac{a}{b-c}=\frac{1}{y}[/tex]
[tex]\frac{c-a}{b}=z \rightarrow \frac{b}{c-a}=\frac{1}{z}[/tex]
-> [tex](\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}). (\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a})=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})[/tex]
Có: [tex](x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=3+\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}[/tex]
Có: [tex]\frac{y+z}{x}=\frac{\frac{b-c}{a}+\frac{c-a}{b}}{\frac{a-b}{c}}=\frac{b^2-bc+ac-a^2}{ab}.\frac{c}{a-b}=\frac{(b-a)(b+a)+c(a-b)}{ab}.\frac{c}{a-b}=\frac{c}{a-b}.\frac{(-a-b+c)(a-b)}{ab}=\frac{2c^2}{ab}[/tex]
tương tự CM được
[tex]\frac{x+z}{y}=\frac{2a^2}{cb}[/tex]
[tex]\frac{x+y}{z}=\frac{2b^2}{ca}[/tex]
Đó, suy ra có: [tex]=3+\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ba}=3+\frac{2a^3}{abc}+\frac{2b^3}{abc}+\frac{2c^3}{bac}=3+2(\frac{a^3+b^3+c^3}{abc})[/tex]
Rồi giờ CM
a^3+b^3+c^3-3abc
=(a+b)^3 -3ab(a+b) +c^3 - 3abc
=[(a+b)^3 +c^3] -3ab.(a+b+c)
=(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)
=(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...
=(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0 (vì a+b+c=0)
[tex]3+2(\frac{a^3+b^3+c^3}{abc})=3+2.\frac{3abc}{abc}=3+2.3=3+6=9(dpcm)[/tex]
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