LT
3
K
Khách
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DH
0
NT
0
HT
0
KM
4
HP
13 tháng 5 2016
\(A=\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}\)
\(A=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10.\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(B=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì 102016+1 < 102017+1
=>\(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
=>\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
=>10A > 10B
=>A > B
13 tháng 5 2016
\(B=\frac{10^{2016}+1}{10^{2017}+1}<\frac{10^{2016}+1+9}{10^{2017}+1+9}\)
\(=\frac{10^{2016}+10}{10^{2017}+10}\)
\(=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}\)
\(=\frac{10^{2015}+1}{10^{2016}+1}=A\)
\(\Rightarrow\) B<A
KT
0
NT
1
6 tháng 5 2017
A=\(\frac{10^{2015}+1}{10^{2016}+1}\)=>10A=\(\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}\)= \(\frac{10^{2016}+10}{10^{2016}+1}\)=\(\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}\)=\(\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}\)=1+\(\frac{9}{10^{2016}+1}\)
B=\(\frac{10^{2016}+1}{10^{2017}+1}\)=>10B=\(\frac{10.\left(10^{2016}+1\right)}{10^{2017+1}}=\frac{10^{2017}+10}{10^{2017}+1}\)= \(\frac{\left(10^{2017}+1\right)+9}{10^{2017}+1}\)=\(\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}\)= 1+\(\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{17}+1\)=>\(\frac{9}{10^{2016}+1}\)>\(\frac{9}{10^{2017}+1}\)nên \(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)=>10A>10B
Vậy A>B
TN
1
Ta có :
\(A=\frac{10^{2016}+1}{10^{2015}+1}=\frac{\left(10^{2016}+1\right).10}{\left(10^{2015}+1\right).10}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10^{2017}+10}{10^{2016}+10}\)
Vì \(10^{2017}=10^{2017}\)và \(10>1\)nên \(10^{2017}+10>10^{2017}+1\)( 1 )
Vì \(10^{2016}=10^{2016}\)và \(10>1\)nên \(10^{2016}+10>10^{2016}+1\)( 2 )
Từ ( 1 ) và ( 2 ) , suy ra : \(\frac{10^{2017}+10}{10^{2016}+10}>\frac{10^{2017}+1}{10^{2016}+1}\)
Vậy \(A>B\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10^{2016}+1+9}{10^{2017}+1+9}=\frac{10^{2016}+10}{10^{2017}+10}=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}=\frac{10^{2015}+1}{10^{2016}+1}\)
lm tương tự vs B ta có
\(A=\frac{10^{2015}+1}{10^{2014}+1}\)
suy ra A>B