so sánh
A=\(\frac{10^{2006}+1}{10^{2007}+1}\)và B=\(\frac{10^{2007}+1}{10^{2008}+1}\)
HEPL ME TO
K
Khách
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PN
1
8 tháng 9 2018
\(A=\frac{10^{2006}+1}{10^{2007}+1}=\frac{10^{2006}+1}{\left(10^{2006}+1\right).10}=\frac{1}{10}\)
\(B=\frac{10^{2007}+1}{10^{2008}+1}=\frac{10^{2007}+1}{\left(10^{2007}+1\right).10}=\frac{1}{10}\)
Nếu đề bài là so sách thì A = B
PK
0
12 tháng 4 2017
\(Tacó:10A=\frac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)\(10B=\frac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)\(Vì:1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
NK
1
5 tháng 2 2016
Đặt A=\(\frac{10^{2006}+1}{10^{2007}+1}\);\(B=\frac{10^{2007}+1}{10^{2008}+1}\)
10A=\(\frac{10\left(10^{2006}+1\right)}{10^{2007}+1}\)=\(\frac{10^{2007}+1+9}{10^{2007}+1}\)
10B=\(\frac{10\left(10^{2007}+1\right)}{10^{2008}+1}=\frac{10^{2008}+1+9}{10^{2008}+1}\)
Vì \(\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}\)nên 10A>10B nên A>B
NN
0
NH
1
12 tháng 12 2015
\(1-A=\frac{10^{2007}-10^{2006}}{10^{2007}+1}=\frac{9.10^{2006}}{10^{2007}+1}=\frac{9.2^{2007}}{10^{2008}+10}\)
\(1-B=\frac{10^{2008}-10^{2007}}{10^{2008}+1}=\frac{9.10^{2007}}{10^{2008}+1}\)
=>1-A< 1-B
=> A > B
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)(\(a;b;m\in\)N*)
Ta có:
\(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)
\(B< \frac{10^{2007}+10}{10^{2008}+10}\)
\(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)
\(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)
=> \(B< A\)
thank you