Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = \(\frac{2013}{2014}+\frac{2014}{2015}=\frac{2014-1}{2014}+\frac{2015-1}{2015}\)
\(=1-\frac{1}{2014}+1-\frac{1}{2015}\)
\(=2-\left(\frac{1}{2014}+\frac{1}{2015}\right)>1\) (1)
b = \(\frac{2013+2014}{2014+2015}<1\) (2)
Từ (1) và (2) => a > b
havsvsuvsvsjzbsvshshsvshjsvdhsjvdhsjdvdhdjdhdhsjdhdhsudghsushdhshshgdgshshdgshdhshdhdghshdgdvshhshdvdgdhshgdgd
h
Ta có: \(A=\frac{2014^{2014}+1}{2014^{2015}+1}\)
\(\Rightarrow2014A=\frac{2014^{2015}+2014}{2014^{2015}+1}=1+\frac{2013}{2014^{2015}+1}\)
\(B=\frac{2014^{2013}+1}{2014^{2014}+1}\)
\(\Rightarrow2014B=\frac{2014^{2014}+2014}{2014^{2014}+1}=1+\frac{2013}{2014^{2014}+1}\)
Mà \(\frac{2013}{2014^{2015}+1}< \frac{2013}{2014^{2014}+1}\Rightarrow1+\frac{2013}{2014^{2015}+1}< \frac{2013}{2014^{2014}+1}\Rightarrow2009A< 2009B\)
\(\Rightarrow A< B\)
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)
Vì \(\frac{10^{2014}+1}{10^{2015}+1}< 1\Rightarrow B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}\)
\(\Rightarrow B< \frac{10^{2014}+10}{10^{2015}+10}\)
\(\Rightarrow B< \frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}\)
\(\Rightarrow B< \frac{10^{2013}+1}{10^{2014}+1}\)
\(\Rightarrow B< A\)
Vậy A > B
Ta có :
\(\frac{2014^{2015}+1}{2014^{2015}+1}\)\(=1\)
\(\frac{2014^{2014}+1}{2014^{2013}+1}\)\(>1\)
\(\Rightarrow A< B\)
Vậy \(A< B\)