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B=(10101+1):(10102+1)<(10101+1+9):(10102 +1+9)=(10101+10):(10102+10)=[10.(10100+1]:[10.(10101+)]
=(10100+1):(10101+1)=A
=>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có A = \(\frac{10^{100}-1}{10^{98}-1}=\frac{10^{98}.10^2-10^2+99}{10^{98}-1}\)
\(=\frac{10^2\left(10^{98}-1\right)+99}{10^{98-1}}\)
\(=10^2+\frac{99}{10^{98}-1}\)
B= \(\frac{10^{101}-1}{10^{99}-1}=\frac{10^{99}.10^2-10^2+99}{10^{99}-1}\)
\(=\frac{10^2\left(10^{99}-1\right)+99}{10^{99}-1}\)
\(=10^2+\frac{99}{10^{99}-1}\)
Vì \(\frac{99}{10^{98}-1}>\frac{99}{10^{99}-1}\)nên \(10^2+\frac{99}{10^{98}-1}>10^2+\frac{99}{10^{99}-1}\)=> A > B
Vậy A > B
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Câu 1 :
Ta có : \(A=\frac{10^{100}+1}{10^{101}+1}\)
\(\Rightarrow10A=\frac{10^{101}+10}{10^{101}+1}=\frac{10^{101}+1+9}{10^{101}+1}=1+\frac{9}{10^{101}+1}\)
Ta có : \(B=\frac{10^{101}+1}{10^{102}+1}\)
\(10B=\frac{10^{102}+10}{10^{102}+1}=\frac{10^{102}+1+9}{10^{102}+1}=1+\frac{9}{10^{102}+1}\)
Vì 10101+1<10102+1
\(\Rightarrow\frac{9}{10^{101}+1}>\frac{9}{10^{102}+1}\)
\(\Rightarrow1+\frac{9}{10^{101}+1}>1+\frac{9}{10^{102}+1}\)
\(\Rightarrow\)10A>10B
\(\Rightarrow\)A>B
Vậy A>B.
Câu 2 :
Ta có : \(E=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì 2001<2001+2002 và 2002<2001+2002
\(\Rightarrow\hept{\begin{cases}\frac{2000}{2001}>\frac{2000}{2001+2002}\\\frac{2001}{2002}>\frac{2001}{2001+2002}\end{cases}}\)
\(\Rightarrow C>E\)
Vậy C>E.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)
Ta có: \(A=\frac{20^{102}+1}{20^{101}+1}< \frac{20^{102}+1+19}{20^{101}+1+19}=\frac{20.\left(20^{101}+1\right)}{20.\left(20^{100}+1\right)}=\frac{20^{101}+1}{20^{100}+1}\)
\(\Rightarrow A< B\)
$\frac{10^{101-1}}{10^{102-1}}$ và $\frac{10^{100+1}}{10^{101+1}}$
= $\frac{10^{100}}{10^{101}}$ và $\frac{10^{101}}{10^{102}}$
Mà $\frac{10^{100}}{10^{101}}$ < $\frac{10^{101}}{10^{102}}$
=> $\frac{10^{101-1}}{10^{102-1}}$ < $\frac{10^{100+1}}{10^{101+1}}$
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