Lời giải:

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

\(\Rightarrow \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\( \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)

\(=\frac{2015.2016c-2015.2017b+2016.2017a-2016.2015c+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)

\(\Rightarrow \left\{\begin{matrix} 2015.2016c-2015.2017b=0\\ 2016.2017a-2016.2015c=0\\ 2017.2015b-2016.2016a=0\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2016c=2017b\\ 2017a=2015c\\ 2015b=2016a\end{matrix}\right.\Rightarrow \frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)

Ta có đpcm.

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, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)

tỉ lệ thức cần chứng minh <=> chứng minh: \(\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\)

Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) = \(\frac{2015a}{2015c}=\frac{2016b}{2016d}=\frac{2016a}{2016c}=\frac{2017b}{2017d}\)

Áp dụng t/c của dãy tỉ số bằng nhau ta có: 

\(\frac{a}{c}=\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\) => đpcm