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Có :
A = 10 - 9/10^1991+1
B = 10 - 9/10^1992+1
Vì 10^1991+1 < 10^1992+1 => 9/10^1991+1 > 9/10^1992+1
=> A < B
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
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A= 10^1992+1/10^1991+1
10/A= 10^1992+1/10^1990+10
=1-9/10^1992+10
B=10^1993+1/10^1993+1
10/B=10^1993+1/10^1993+10
=1-9/10^1993+10
Vi 9/10^99+10>9/10^1993+10
nen A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy \(10^{1993}+1>10^{1992}+1\)
\(\Rightarrow B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10.\left(10^{1992}+1\right)}{10.\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
\(\Rightarrow A< B\)
\(\frac{10^{1993}+1}{10^{1992}+1}>1\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}\)
\(\frac{10^{1993+1}}{10^{1992}+1}>\frac{10^{1993}+10}{10^{1992}+10}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}+1}{10^{1991}+1}\)
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=1994
=1992-1991-1993=1-1993=-1992