Question

cho \(\dfrac{a}{2003}=\dfrac{b}{2005}=\dfrac{c}{2007}\)

CMR :\(\dfrac{\left(a-c\right)^2}{4}\)= (a-b).(b-c)

Answer 1

Đặt \(\dfrac{a}{2003}=\dfrac{b}{2005}=\dfrac{c}{2007}=k\Rightarrow\left\{{}\begin{matrix}a=2003k\\b=2005k\\c=2007k\end{matrix}\right.\)

Ta có: \(\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(2003k-2007k\right)^2}{4}=\dfrac{16k^2}{4}=4k^2\) (1)

\(\left(a-b\right)\left(b-c\right)=\left(2003k-2005k\right)\left(2005k-2007k\right)\)

\(=2k2k=4k^2\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\left(a-b\right)\left(b-c\right)\left(đpcm\right)\)

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