$a+b+c=2007\Rightarrow \frac{1}{a+b+c}=\frac{1}{2007}\Rightarrow \frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow \frac{1}{a+b+c}-\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\Rightarrow \frac{-(b+c)}{a(a+b+c)}=\frac{b+c}{bc}\Rightarrow (b+c)\left [ \frac{1}{a(a+b+c)}+\frac{1}{bc} \right ]=0\Leftrightarrow (b+c)\left [ \frac{a^2+ab+ac+bc}{abc(a+b+c)} \right ]=0\Leftrightarrow (a+b)(b+c)(a+c)\left [ \frac{1}{abc(a+b+c)} \right]\Rightarrow (a+b)(b+c)(a+c)=0\Rightarrow \begin{bmatrix} a+b=0\\b+c=0 \\ c+a=0 \end{bmatrix}$
mà a+b+c=2007 suy ra một trong ba số bằng 2007
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- perfectstrong, tpdtthltvp và bovuotdaiduong thích