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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)
