So sánh A và B biết:
A = 20002016 + 20002017 ; B = 20012017
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,33^{44}=11^{44}\cdot3^{44}=11^{44}\cdot81^{11}>11^{33}\cdot64^{11}=11^{33}\cdot4^{33}=44^{33}>44^{32}\)
\(b,A=2000^{2016}\left(2000-1\right)+1999=1999\cdot2000^{2016}+1999⋮1999\)
A=1-1/(2013*2014)
B=1-1/(2014*2015)
2013*2014<2014*2015
=>1/2013*2014>1/2014*2015
=>-1/2013*2014<-1/2014*2015
=>A<B
A=2011^2012-2011^2011= 2011^2011 * 2011 -2011^2011= 2011^2011 *(2011-1)= 2011^2011 *2010
B=2011^2013-2011^2012=2011^2012*2011- 2011^2012= 2011^2012 *(2011-1) = 2011^2012 *2010
vì 2011^2011*2010 < 2011^2012*2010 nên A<B
Ta có : 2011^2013 x M = (2010^2012 x 2011 + 2011^2013)^2013 > (2010^2013 + 2011^2013)^2013 = N x (2010^2013 + 2011^2013)
Do đó: 2011^2013 x M > N x (2010^2013 + 2011^2013)
<=> M > N x [(2010/2011)^2013 + 1] ==> M > N (điều phải chứng minh)
10A=10*\(\frac{10^{2006}+1}{10^{2007}+1}\) 10B=10*\(\frac{10^{2007}+1}{10^{2008}+1}\)
10A=\(\frac{10^{2007}+1+9}{10^{2007}+1}\) 10B=\(\frac{10^{2008}+1+9}{10^{2008}+1}\)
10A=1+\(\frac{9}{10^{2007}+1}\) 10B=1+\(\frac{9}{10^{2008}+1}\)
Vì \(\frac{9}{10^{2007}+1}\)>\(\frac{9}{10^{2008}+1}\)=>1+\(\frac{9}{10^{2007}+1}\)>1+\(\frac{9}{10^{2008}+1}\)
Nên 10A>10B=>A>B
Ta có: \(A=\frac{10^{2006}+1}{10^{2007}+1}\)
\(=>10A=\frac{10^{2007}+10}{10^{2007}+1}=\frac{10^{2007}+1+9}{10^{2007}+1}=\frac{10^{2007}+1}{10^{2007}+1}+\frac{9}{10^{2007}+1}=1+\frac{9}{10^{2007}+1}\)
\(B=\frac{10^{2007}+1}{10^{2008}+1}\)
\(=>10B=\frac{10^{2008}+10}{10^{2008}+1}=\frac{10^{2008}+1+9}{10^{2008}+1}=\frac{10^{2008}+1}{10^{2008}+1}+\frac{9}{10^{2008}+1}=1+\frac{9}{10^{2008}+1}\)
Vì \(10^{2007}+1< 10^{2008}+1=>\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}=>1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}=>10A>10B=>A>B\)
Ta có : A = 20002016 + 20002017
= 20002016.(1 + 2000)
= 20002016.2001
< 20012016.2001
= 20012017 = B
=> A < B
Vậy A < B
B=20002017+2017 ,A=20002016+20002017
Mà 20002016>2017
=>A>B