A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)

B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)

Rồi bạn tự so sánh nha

A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)

\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)

\(Vậy:A>B\)

Đúng nha Nguyễn Bình Minh

so sánh:

\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)  và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)

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

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)

          \(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)

          \(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)

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

Vậy: \(A>B\)

gọi  \(A=\frac{2015^{2015}+1}{2015^{2016}+1};B=\frac{2015^{2014}+1}{2015^{2015}+1}\)

         \(\Rightarrow A=\frac{2015^{2015}+1}{2015^{2016}+1}<\frac{2015^{2015}+2014+1}{2015^{2016}+2014+1}=\frac{2015^{2015}+2015}{2015^{2016}+2015}=\frac{2015\left(2015^{2014}+1\right)}{2015\left(2015^{2015}+1\right)}=\frac{2015^{2014}+1}{2015^{2015}+1}=B\)

so sánh: \(A=\frac{2014}{2015}+\frac{2015}{2016}\)  và \(B=\frac{2014+2015}{2015+2016}\)

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

Ta có: \(\frac{2014}{2015}>\frac{2014}{2015+2016}\) vì \(2015<2015+2016\)

          \(\frac{2015}{2016}>\frac{2015}{2015+2016}\) vì \(2016<2015+2016\)

\(\Rightarrow\frac{2014}{2015}+\frac{2015}{2016}>\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)

\(\Rightarrow\frac{2014}{2015}+\frac{2015}{2016}>\frac{2014+2015}{2015+2016}\)

Vậy:     \(A>B\)

A = (n + 2015)(n + 2016) + n² + n

= (n + 2015)(n + 2015 + 1) + n(n + 1)

Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2

=> (n + 2015)(n + 2015 + 1) chia hết cho 2

      n(n + 1) chia hết cho 2

=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2

=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)

cac ban sai het ruk do. phai la 70 moi dung.

xin loi. an nham