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A=10^2015-1 / 10^2016-1 và B= 10^2014+1 / 10^2015+1
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A- 1 = \(\frac{10^{2015}-1-\left(10^{2016}-1\right)}{10^{2016}-1}=\frac{-9.10^{2015}}{10^{2016}-1}=\frac{-90.10^{2014}}{10^{2016}-1};\)
B- 1 = \(\frac{10^{2014}+1-\left(10^{2015}+1\right)}{10^{2015}+1}=\frac{-9.10^{2014}}{10^{2015}+1};\)
xét \(\frac{A-1}{B-1}=\frac{-90.10^{2014}}{10^{2016}-1}:\frac{-9.10^{2014}}{10^{2015}+1}=\frac{10\left(10^{2015}+1\right)}{10^{2016}-1}=\frac{10^{2016}+10}{10^{2016}-1}>1\)
=> A-1 > B-1 => A > B
\(10A=\dfrac{10^{2015}+2016+9\cdot2016}{10^{2015}+2016}=1+\dfrac{18144}{10^{2015}+2016}\)
\(10B=\dfrac{10^{2016}+9+18144}{10^{2016}+2016}=1+\dfrac{18144}{10^{2016}+2016}\)
mà \(\dfrac{18144}{10^{2015}+2016}>\dfrac{18144}{10^{2016}+2016}\)
nên A>B
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
Ta có: \(10A=10.\left(\frac{10^{2014}+1}{10^{2015}+1}\right)=\frac{10^{2015}+10}{10^{2015}+1}=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)
\(10B=10.\left(\frac{10^{2015}+1}{10^{2016}+1}\right)=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
Vì 1 = 1; 9 = 9 ta so sánh mẫu:
Ta có: 102015 < 102016 => 102015+1 < 102016+1
=> \(1+\frac{9}{10^{2015}+1}>1+\frac{9}{10^{2016}+1}\)
=> 10A > 10B
=> A > B.
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
Ta có:
\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)
\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)
Vì \(10^{2016}+2020>2^{2015}+2020\)
=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)
=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)
=> 10B < 10A
=> B<A