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

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

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

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

=>đpcm

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

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

a,a=b+1

suy ra a-b=1 suy ra(\(\sqrt{a}+\sqrt{b}\))(\(\sqrt{a}-\sqrt{b}\))=1

suy ra \(\sqrt{a}-\sqrt{b}\)=\(\frac{1}{\sqrt{a}+\sqrt{b}}\)(1)

vì a=b+1 suy ra a>b suy ra \(\sqrt{a}>\sqrt{b}\)suy ra \(\sqrt{a}+\sqrt{b}>2\sqrt{b}\)

suy ra \(\frac{1}{\sqrt{a}+\sqrt{b}}< \frac{1}{2\sqrt{b}}\)(2)

từ (1) ,(2) suy ra\(\sqrt{a}-\sqrt{b}< \frac{1}{2\sqrt{b}}\)suy ra \(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}\)(*)

ta lại có b+1=c+2 suy ra b-c =1 suy ra\(\left(\sqrt{b}-\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)=1\)

suy ra \(\sqrt{b}-\sqrt{c}=\frac{1}{\sqrt{b}+\sqrt{c}}\)(3)

vì b>c suy ra \(\sqrt{b}>\sqrt{c}\) suy ra \(\sqrt{b}+\sqrt{c}>2\sqrt{c}\)

suy ra \(\frac{1}{\sqrt{b}+\sqrt{c}}< \frac{1}{2\sqrt{c}}\)(4)

Từ (3),(4) suy ra \(\sqrt{b}-\sqrt{c}< \frac{1}{2\sqrt{c}}\) suy ra\(2\left(\sqrt{b}+\sqrt{c}\right)< \frac{1}{\sqrt{c}}\)(**)

từ (*),(**) suy ra đccm