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)

a)\(\frac{2013}{2015}< \frac{2014}{2016}\)

b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)

ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........

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: \(A>B\)

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015}\) (1)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\) (2)
ộng caác bất đẳng thứa (1) và (2) vào vế với vế:
\(\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\Rightarrow A>B\)

cảm ơn các bạn                         

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

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

\(=6048>2\)

Vậy: \(C>D\)

\(B-1=\frac{2015^{2014}+1}{2015^{2013}+1}-1=\frac{2015^{2015}+2015}{2015^{2014}+2015}-1=\frac{2015^{2015}-2015^{2014}}{2015^{2014}+2015}\)

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

=> A- 1 > B- 1 => A>B

Câu b) Làm tương tự bạn nhé

2 cái bang nhau

= chắc chắn 100%

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mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15