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

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

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

=>đpcm

đặt a/2014=b/2015=c/2016=k

=>a=2014k;b=2015k;c=2016k

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

=4.k(2014-2015).k92015-2016)=4.k.(-1).k.(-1)=4.k^2(1)

=>(c-a)(c-a)=(c-a)^2=(2016k-2014k)(2016k-2014k)=[k(2016-2014)]^2=[k.2]^2=k^2.4(2)

từ (1)và (2)=>4(a-b)(b-c) = (c-a).(c-a)

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

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

Lời giải:

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

$\Rightarrow a=2014k; b=2015k; c=2016k$

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

$=4(-k)(-k)=4k^2(1)$

Và:

$(c-a)^2=(2016k-2014k)^2=(2k)^2=4k^2(2)$

Từ $(1); (2)\Rightarrow 4(a-b)(b-c)=(c-a)^2$ (đpcm)

Đặt a/2014 = b/2015 = c/2016 = k => a = 2014k; b = 2015k; c= 2016k

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

        =4(-1k)(-1k)=4k^2 (1) (c-a)^2

        =(2016-2014)^2=(2k)^2=4k^2 (2)

Từ (1) và (2) => ............

1; 2 ;3

Đặt dãy tỉ số = k 

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

Thay a,b,c vào đẳng thức dưới => ĐPCM