Đặt \(\frac{a}{2013}=\frac{b}{2014}=\frac{c}{2015}=k\) => a=2013k; b=2014k; c=2015k

Ta có: 4(a-b)(b-c) = 4(2013k-2014k)(2014k-2015k)

= 4(-k)(-k) = 4k² (1)

Lại có: (c-a)² = (2015k-2013k)² = (2k)² = 4k² (2)

Từ (1) và (2) => 4(a-b)(b-c)=(c-a)² (đpcm)

Bài 2)

Ta có \(\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)

Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Rightarrow ab+ad< ab+bc\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)

Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)

\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Rightarrow ad+cd< bc+cd\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)

Đặt \(\frac{a}{2013}\) = \(\frac{b}{2014}\) = \(\frac{c}{2015}\) = k

nên a = 2013k; b = 2014k và c = 2015k

Xét hiệu:

4(2013k - 2014k)(2014k - 2015k) - (2015k - 2013k)²

= 4(-k)(-k) - (2k)²

= 4k² - 4k² = 0

Vậy 4(a - b)(b - c) = (c - a)².

Ta có : \(\frac{a+2014}{a-2014}=\frac{a+2015}{a-2015}\)

\(\Rightarrow\left(a+2014\right)\left(a-2015\right)=\left(a-2014\right)\left(a+2015\right)\)

\(\Rightarrow a^2-a-2014.2015=a^2+a-2014.2015\)

\(\Leftrightarrow a^2-a=a^2+a\)

=> a² - a² - a = a

=> -a = a

=>  0 = a + a

=> 2a = 0

=> a = 0 

Vậy \(\frac{a}{2014}=\frac{b}{2015}\) (đpcm)

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

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

\(TA-CO':\)

\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)

\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(A=\frac{4}{7}\)

\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)

ĐẶT \(C=1+2+...+2^{2013}\)

\(\Rightarrow2C=2+2^2+...+2^{2014}\)

\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)

\(\Rightarrow C=2^{2014}-2\)

\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)

\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)

\(B=\frac{1}{2}\)

\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)

\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)

VẬY, \(A-B=\frac{-1}{14}\)

bài gì khó thế!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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