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1. \(G=2016.2016=\left(2014+2\right)\left(2018-2\right)=2014.2018-4028+4036-4=2014.2018+4\)
vì 2014.2018+4 >2014.2018
=> G>H
\(\frac{2016.2016}{2013.2019}=\frac{\left(2013+3\right)\left(2019-3\right)}{2013.2019}=\frac{2013.2019-6039+6057-9}{2013.2019}=\frac{2013.2019+9}{2013.2019}=1+\frac{9}{2013.2019}\)
vì \(1+\frac{9}{2013.2019}>1\)
\(\frac{2016.2016}{2013.2019}>1\)
A = 2016^2015 +1 / 2016^2014+1 < 2016^2015 + 1 + 2015 / 2016^2014 + 1 + 2015
= 2016^2015 + 2016 / 2016^2014 + 2016
= 2016(2016^2014 + 1 ) / 2016(2016^2013 +1)
= 2016^2014 + 1 / 2016^2013 + 1 = B
=> A < B
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: 58/63=3190/3465
36/55=2268/3465
=>58/63>36/55
b: 27/53=1998/3922
36/74=1908/3922
=>27/53>36/74
Xét \(A=\frac{10^{2014}+2016}{10^{2015}+2016}\Rightarrow10A=\frac{10^{2015}+20160}{10^{2015}+2016}=\frac{10^{2015}+2016+18144}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\)
Xét \(B=\frac{ 10^{2015}+2016}{10^{2016}+2016}\Rightarrow10B=\frac{10^{2016}+20160}{10^{2016}+2016}=\frac{10^{2016}+2016+18144}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2016}\)
Có \(\frac{18144}{10^{2015}+2016}>\frac{18144}{10^{2016}+2016}\)
\(\Rightarrow10A>10B\Leftrightarrow A>B\)
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)