so sanh A B biet A=\(\frac{10^{2018}+1}{10^{2017}+1}\);B=\(\frac{10^{2007}+1}{10^{2008+1}}\)
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24 tháng 7 2018
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)(1)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)(2)
Từ(1) và (2)
\(\Rightarrow B>A\)
QT
0
13 tháng 12 2018
Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :
\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)
\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1;9=9\)
\(\Rightarrow\)Ta so sánh mẫu , ta có:
\(10^{2017}< 10^{2018}\)
\(\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
Hay \(A>B\)
TG
3
T
19 tháng 10 2018
\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)
\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow B=2^{2019}-1\)
\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)
LH
3
21 tháng 4 2018
Ta có: \(\hept{\begin{cases}A=\frac{10^{2016}+1}{10^{2017}+1}\\B=\frac{10^{2017}+1}{10^{2018}+1}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}10A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\\10B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\end{cases}}\)
Vì \(\frac{9}{10^{1017}+1}>\frac{9}{10^{2018}+1}\)
nên \(10A>10B\Rightarrow A>B\)
21 tháng 4 2018
\(A=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\frac{10\cdot(10^{2016}+1)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(A=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{2017}+1\)
\(\Rightarrow\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
\(\Rightarrow\)\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
....
KV
18 tháng 3 2019
ta có :
\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)
\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)
Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)
\(\Rightarrow A< B\)
Ta có: \(10^{2018}>10^{2017}\Rightarrow10^{2018}+1>10^{2017}+1\Rightarrow A=\frac{10^{2018}+1}{10^{2017}+1}>1\) (1)
\(10^{2007}< 10^{2008}\Rightarrow10^{2007}+1< 10^{2008}+1\Rightarrow B=\frac{10^{2007}+1}{10^{2008}+1}< 1\) (2)
Từ (1) và (2) => A > B