\(N=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)

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

\(N=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)

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

\(\Rightarrow\frac{M}{N}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}=\frac{1}{2017}\)

hay lam

Ta có : \(\left(x-2\right)^{2016}\)dương

\(\Rightarrow x-3=0\Rightarrow x=3\)

Thay x ta thử :

 \(\left(3-2\right)^{2016}+\left(3-3\right)=1+0=1\)thỏa đề

Vậy \(x=3\)

bài 2)

ta có 

= 2015 +2015^2+2015^3+2015^4+2015^5+2015^6 

= (2015 +2015^2)+(2015^3+2015^4)+(2015^5+2015^6)

= (2015.1+2015.2015)+ ... +(2015^5.1+2015^5.2015)

= 2015.2016+...+2015^5.2016

= 2016.(2015+2015^3+2015^5) chia hết cho 2016

=> (2015 +2015^2+2015^3+2015^4+2015^5+2015^6) chia het cho 2016

\(\frac{n-4}{2016}+\frac{n-3}{2015}=\frac{n-2}{2014}+\frac{n-1}{2013}\)

\(\Rightarrow\left(\frac{n-4}{2016}+1\right)+\left(\frac{n-3}{2015}+1\right)=\left(\frac{n-2}{2014}+1\right)+\left(\frac{n-1}{2013}+1\right)\)

\(\Rightarrow\frac{n-4+2016}{2016}+\frac{n-3+2015}{2015}=\frac{n-2+2014}{2014}+\frac{n-1+2013}{2013}\)

\(\Rightarrow\frac{n+2013}{2016}+\frac{n+2013}{2015}=\frac{n+2013}{2014}+\frac{n+2013}{2013}\)

\(\Rightarrow\frac{n+2013}{2016}+\frac{n+2013}{2015}-\frac{n+2013}{2014}-\frac{n+2013}{2013}=0\)

\(\Rightarrow\left(n+2013\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)

Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\ne0\)

=> n + 2013 = 0 => n = -2013

Vậy n = -2013

bạn ơi,cách giải của bạn đúng rồi nhưng n-4+2016=n+2012 , mấy số kia cũng thế chứ ạ

-2013 nha