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\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
Ta có:
\(\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)
\(=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)
\(=1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2016}+\dfrac{2017}{2017}\)
\(=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
Do đó: \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\right)}=\dfrac{1}{2017}\)
Vậy...
\(\dfrac{a+b+c}{b+c-2016+a+c+2015+a+b+1}=\dfrac{a+b+c}{2016}\)
Sau đó rút gọn đi rồi tính nha
STUDY WELL 
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b+c}{b+c-2016+a+c+2015+a+b+1}=\dfrac{a+b+c}{2016}\\ \Rightarrow\dfrac{a+b+c}{2a+2b+2c}=\dfrac{a+b+c}{2016}\\ \Rightarrow\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2016}\)
\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
Ta có:
*) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(\Rightarrow S=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Vậy \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)
\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)
\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)
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.