NK
2
K
Khách
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NN
1
NB
6 tháng 3 2022
\(10A=10.\dfrac{10^{2004}+1}{10^{2005}+1}=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\\ 10B=10.\dfrac{10^{2005}+1}{10^{2006}+1}=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)
vì \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\Rightarrow10A>10B\Rightarrow A>B\)
M
1
20 tháng 5 2021
Giải:
Ta có:
A=\(\dfrac{10^{2019}-1}{10^{2020}+1}\)
10A=\(\dfrac{10^{2020}-10}{10^{2020}+1}\)
10A=\(\dfrac{10^{2020}+1-11}{10^{2020}+1}\)
10A=\(1+\dfrac{-11}{10^{2020}+1}\)
Tương tự:
B=\(\dfrac{10^{2020}-1}{20^{2021}+1}\)
10B=\(1+\dfrac{-11}{10^{2021}+1}\)
Vì \(\dfrac{-11}{10^{2020}+1}< \dfrac{-11}{10^{2021}+1}\) nên 10A<10B
⇒A<B
Chúc bạn học tốt!
NA
2
25 tháng 9 2021
\(A=1+2+2^2+2^3+...+2^{2021}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2022}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}=2^{2022}-1>2^{2021}-1=N\)
25 tháng 9 2021
\(a=1+2+2^2+...+2^{2021}\\ \Rightarrow2a=2+2^2+2^3+...+2^{2022}\\ \Rightarrow2a-a=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow a=2^{2022}-1>2^{2021}-1=n\)
KT
3
giúp mk vs
NT
0
LT
1
TH
29 tháng 8 2018
Ta có:
\(A=\dfrac{2010^{2011}+1}{2010^{2012}+1}\)
\(A< \dfrac{2010^{2011}+1+2009}{2010^{2012}+1+2009}\)
\(A< \dfrac{2010^{2011}+2010}{2010^{2012}+2010}\)
\(A< \dfrac{2010\left(2010^{2010}+1\right)}{2010\left(2010^{2011}+1\right)}\)
\(A< \dfrac{2010^{2010}+1}{2010^{2011}+1}\)
Mà \(B=\dfrac{2010^{2010}+1}{2010^{2011}+1}\)
\(\Rightarrow A< B\)
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Rightarrow A>B\)
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Theo đề ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Lúc này ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Leftrightarrow A>B\).