Ta có: \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=\frac{a-b}{2009-2010}=\frac{b-c}{2010-2011}=\frac{c-a}{2011-2009}.\)
\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Rightarrow\frac{a-b}{-1}.\frac{b-c}{-1}=\left(\frac{c-a}{2}\right)^2\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{2^2}\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{4}.\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2.1\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)-\left(c-a\right)^2=0.\)
Hay \(M=0.\)
Vậy \(M=0.\)
Chúc bạn học tốt!
Đặt \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=k\)
\(=>\hept{\begin{cases}a=2009k\\b=2010k\\c=2011k\end{cases}}\)
Ta có : \(4\left(a-b\right)\left(b-c\right)=4\left(2009k-2010k\right)\left(2010k-2011k\right)\)
\(=4\left(-k\right)\left(-k\right)=4k^2\)
Lại có : \(\left(c-a\right)^2=\left(2011k-2009k\right)^2=\left(2k\right)^2=4k^2\)
Suy ra \(M=4k^2-4k^2=0\)