So sánh a=10^2006+1/10^2007+1 VÀ B=10^2007+1/10^2008+1
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TT
4
16 tháng 5 2016
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
16 tháng 5 2016
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 \(10^{2007}+1< 10^{2008}+1=>\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}=>1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}=>10A>10B=>A>B\)
NV
1
PV
2
7 tháng 8 2016
Ta có: A=\(\frac{10^{2006}+1}{10^{2007}+1}\)
=>10A=\(\frac{10\left(10^{2006}+1\right)}{10^{2007}+1}=\frac{10^{2007}+10}{10^{2007}+1}=1+\frac{9}{10^{2007}+1}\)
Ta có: B=\(\frac{10^{2007}+1}{10^{2008}+1}\)
=>10B=\(\frac{10\left(10^{2007}+1\right)}{10^{2008}+1}=\frac{10^{2008}+10}{10^{2008}+1}=1+\frac{9}{10^{2008}+1}\)
Mà \(\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}\) (do 102007+1<102008+1)
=>\(1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}\)
=>10A>10B
=>A>B
7 tháng 8 2016
Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
=> \(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)
=> \(B< \frac{10^{2007}+10}{10^{2008}+10}\)
=> \(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)
=> \(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)
NL
1
D
1
TN
14 tháng 6 2016
A=\(\frac{10^{2006}+1}{10^{2007}+1}\)
10.A=\(\frac{10.\left(10^{2006}+10\right)}{10^{2007}+1}\)
=\(1+\frac{9}{10^{2007}+1}\)
B=\(\frac{10^{2007}+1}{10^{2008}+1}\)
\(10.B=\frac{10.\left(10^{2007}+10\right)}{10^{2008}+1}\)
= \(1+\frac{9}{10^{2008}+1}\)
Vì\(1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}\) nên 10A > 10B \(\Rightarrow A>B\)
k cko mk nka
10 tháng 9 2016
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)(\(a;b;m\in\)N*)
Ta có:
\(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)
\(B< \frac{10^{2007}+10}{10^{2008}+10}\)
\(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)
\(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)
=> \(B< A\)