\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=\frac{a-b}{2014-2015}=\frac{b-c}{2015-2016}=\frac{c-a}{2016-2014}\)

=\(\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)=>\(\frac{\left(a-b\right)\left(b-c\right)}{\left(-1\right)\left(-1\right)}=\frac{\left(c-a\right)^2}{2^2}=\frac{\left(a-b\right)\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{4}\Leftrightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Gọi \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\Rightarrow a=2014k;b=2015k;c=2016k\left(1\right)\)

Thay (1) vào M ta có :

M=4(2014k-2015k)(2015k-2016k)-(2016k-2014k)²

=>M=4.-k.-k-4k²

=>M=4k²-4k²=0

Vậy M = 0

đặt \(\frac{a}{2014}\)=\(\frac{b}{2015}\)=\(\frac{c}{2016}\)= K

---> a = 2014k, b=2015k , c=2016k

về trái : 4. ( 2014k-2015k). (2015k-2016k)=4. (-1k).(-1k)=4k²

Về phai: (2016k-2014k)²=(2k)²=4k²

---> ve trai = ve phai----> dpcm

Đặt : \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)

\(\Rightarrow\frac{a}{2014}=k\Rightarrow a=2014k\)

\(\Rightarrow\frac{b}{2015}=k\Rightarrow b=2015k\)

\(\Rightarrow\frac{c}{2016}=k\Rightarrow c=2016k\)

Ta có : \(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)\)

\(=4k\left(2014-2015\right).k\left(2015-2016\right)=4k.\left(-1\right).k.\left(-1\right)=4.k^2\)( 1 )

\(\Rightarrow\left(c-a\right)^2=\left(2016k-2014k\right)\left(2016k-2014k\right)=\left[\left(2016k-2014k\right)^2\right]=\left[k\left(2016-2014\right)\right]=\left(k^2\right)^2=k^{2.4}\)( 2 )

Từ \(\left(1\right)\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)

\(\Rightarrow a=2014k;b=2015k;c=2016k\)

\(\Rightarrow4(a-b)(b-c)=4(2014k-2015k)(2015k-2016k)\)

\(\Rightarrow4\cdot k(2014-2015)\cdot k(2015-2016)=4\cdot k\cdot(-1)\cdot k\cdot(-1)=4\cdot k^2\)

\(\Rightarrow(c-a)(c-a)=(c-a)^2=(2016k-2014k)=[k(2016-2014)]^2=(k\cdot2)^2=k^{2\cdot4}\)

Rồi tự suy ra đấy

Bạn Namikaze Minato làm đúng rồi đấy

\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=\frac{a-b}{2014-2015}\)

\(=\frac{b-c}{2015-2016}=\frac{c-a}{2016-2014}\)

\(=\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)

\(\Rightarrow a-b=-\frac{c-a}{2};b-c=-\frac{c-a}{2}\)

do đó: \(\left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}\)

\(\Rightarrow M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2=0\)

Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)

=> \(\hept{\begin{cases}a=2014k\\b=2015k\\c=2016k\end{cases}}\)

Suy ra \(M=4\left(2014k-2015k\right)\left(2015k-2016k\right)-\left(2016k-2014k\right)^2=4k^2-4k^2=0\)

Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\Rightarrow a=2014k;b=2015k;c=2016k\)

=>\(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)

\(\left(c-a\right)^2=\left(2016k-2014k\right)^2=\left(2k\right)^2=4k^2\)

=>đpcm