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

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 dãy tỉ số = k => a = 2014k , b = 2015k , c = 2016k Thay a,b,c vào đẳng thức dưới => ĐPCM 

Nhớ mặt từ sau đừng bảo tui giải cho

Ta có :\(\frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}=\frac{b-a}{2016-2015}=\frac{c-a}{2017-2015}=\frac{c-b}{2017-2016}\)

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

\(\Rightarrow\hept{\begin{cases}2\left(b-a\right)=c-a\\2\left(b-c\right)=c-a\end{cases}\Rightarrow4\left(b-a\right)\left(b-c\right)=\left(c-a\right)^2}\)

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

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

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)\)

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

\(=4.\left(-k\right).\left(-k\right)\)

\(=4k^2\)(1)

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

Từ (1) và (2) 

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

                                                    đpcm

Tham khảo nhé~