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: \(A>B\)

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)

\(C=\dfrac{2013}{2013}+2014+\dfrac{2014}{2014}+2015+\dfrac{2015}{2015}+2016\)

\(=1+2014+1+2015+1+2016\)

\(=6048>2\)

Vậy: \(C>D\)

de ot la dau = nha

\(\frac{2016}{2015}>1;\frac{2015}{2014}>1;\frac{2014}{2013}>1.\)

\(\Rightarrow\frac{2016}{2015}+\frac{2015}{2014}+\frac{2014}{2013}>1+1+1=3\)

Vậy A>3