Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Đặt :
\(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=k\)
=> a=2009k
b=2011k
c=2013k
Thay vào ta có :
\(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
hay
\(\frac{\left(2009k-2013k\right)^2}{4}=\left(2009k-2013k\right)\left(2011k-2013k\right)\)
\(\Rightarrow\frac{\left(-4k\right)^2}{4}=\left(-4k\right)\left(-2k\right)\)
\(\Rightarrow\frac{\left(-4k\right)^2}{4}=8k^2\)
Ta lại có :
\(\left(-4k\right)^2=\left(-4.-4\right).\left(k^2\right)\)
Thay vào ta được :
\(\frac{\left(16.k^2\right)}{4}=8.k^2\)
Mà \(\frac{\left(16.k^2\right)}{4}=8.k^2\)
(đ.p.c.m)
Nguyễn Đình Dũng
ntn hả Nguyễn Huy Thắng
