\(\frac{a}{2003}=\frac{b}{2004}=\frac{a-b}{2003-2004}=-\left(a-b\right)\) = -(b-c)=\(\frac{c-a}{2}\)
=> -(a-b).(-(b-c)=\(\frac{c-a}{2}.\frac{c-a}{2}=\frac{\left(c-a\right)^2}{4}\)
<=> 4.(a-b).(b-c)=(c-a)2
Đặt \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\left(k\ne0\right)\)
\(\Rightarrow a=2003k\), \(b=2004k\), \(c=2005k\)
Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2003k-2004k\right)\left(2004k-2005k\right)\)
\(=4.\left(-k\right).\left(-k\right)=4k^2\)(1)
Mặt khác ta có: \(\left(c-a\right)^2=\left(2005k-2003k\right)^2=\left(2k\right)^2=4k^2\)(2)
Từ (1) và (2) \(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)( đpcm )
Đặt \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\Rightarrow\hept{\begin{cases}a=2003k\\b=2004k\\c=2005k\end{cases}}\)
*\(4\left(a-b\right)\left(b-c\right)=4\left(2003k-2004k\right)\left(2004k-2005k\right)\)
\(=4\left(-k\right)\left(-k\right)=4k^2\)(1)
*\(\left(c-a\right)^2=\left(2005k-2003k\right)^2=\left(2k\right)^2=4k^2\)(2)
Từ (1) và (2) => đpcm
