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a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
\(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\Rightarrow\left(a+2014\right)\left(b-2015\right)=\left(a-2014\right)\left(b+2015\right)\)
\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a-2014}{b-2015}=\frac{a+2014+a-2014}{b+2015+b-2015}=\frac{2a}{2b}=\frac{a}{b}\)
\(\Rightarrow\frac{a+2014}{b+2015}=\frac{a}{b}=\frac{a+2014-a}{b+2015-b}=\frac{2014}{2015}\)
\(\frac{a}{b}=\frac{2014}{2015}\Rightarrow2015a=2014b\Rightarrow\frac{a}{2014}=\frac{b}{2015}\)
\(\Rightarrowđpcm\)
\(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\Rightarrow\left(a+2014\right)\left(b-2015\right)=\left(a-2014\right)\left(b+2015\right)\)
\(\Rightarrow\) \(ab+2014b-2015a-2014.2015=ab+2015a-2014b-2014.2015\)
\(\Rightarrow\) \(\left(ab-ab\right)+\left(-2014.2015+2014.2015\right)=\left(2015a+2015a\right)-\left(2014b+2014b\right)\)
\(\Rightarrow0+0=4030a-4028b\)
\(\Rightarrow4030a=4028b\) \(\Rightarrow\frac{a}{b}=\frac{4028}{4030}=\frac{2014}{2015}\Rightarrow\frac{a}{2014}=\frac{b}{2015}\)
Vậy nếu \(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\) thì \(\frac{a}{2014}=\frac{b}{2015}\) (đpcm)
Ta có :
\(\frac{a}{b}< \frac{2015}{2013}\)
\(\Rightarrow2013a< 2015b\)
\(\Rightarrow2013a+ab=2015b+ab\)
\(\Rightarrow a.\left(2013+b\right)=b.\left(2015+a\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a+2015}{b+2013}\)
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)
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)
Các bạn k cần trả lời nữa! Thông cảm nha! 
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)