Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Ta có: \(\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2019}-1+11}{10^{2020}-1+11}=\frac{10^{2019}+10}{10^{2020}+10}=\frac{10.\left(10^{2018}+1\right)}{10.\left(10^{2019}+1\right)}=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2018}+1}{10^{2019}+1}\)
Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)
\(B=\frac{10^{2018}+1}{10^{2019}+1}\)
Dễ thấy \(A< 1\)
Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với m>0
Vậy \(A=\frac{10^{2019}-1}{10^{2020}-1}< \frac{\left[10^{2019}-1\right]+11}{\left[10^{2020}-1\right]+11}=\frac{10^{2019}+10}{10^{2020}+10}\)
\(A< \frac{10\left[10^{2018}+1\right]}{10\left[10^{2019}+1\right]}=\frac{10^{2018}+1}{10^{2019}+1}=B\)
Do đó : A<B
#)Giải :
Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)
\(B=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow10A=\frac{10^{2020}-10}{10^{2020}-1}=\frac{10^{2020}-1+9}{10^{2020}-1}=\frac{10^{2020}-1}{10^{2020}-1}+\frac{9}{10^{2020}-1}=1+\frac{9}{10^{2020}-1}\)
\(\Rightarrow10B=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=\frac{10^{2019}+1}{10^{2019}+1}+\frac{9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
Dễ thấy \(10^{2020}-1>10^{2019}+1\)
\(\Rightarrow\frac{9}{10^{2020}-1}>\frac{9}{10^{2019}+1}\Rightarrow1+\frac{9}{10^{2020}-1}>1+\frac{9}{10^{2019}+1}\Rightarrow10A>10B\Rightarrow A>B\)
\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}>\frac{10^{2018}+1}{10^{2019}+1}\)
