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\(A=\frac{10^{2017}}{10^{2018+1}}=\frac{10^{2017}}{10^{2019}}=\frac{1}{10^2}\)
Tương Tự với \(B=\frac{1}{10^2}\)
\(\Rightarrow A=B\)
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
Có: \(A=\frac{10^{2017}}{10^{2018}+1}\)
\(\Rightarrow10A=\frac{10^{2018}}{10^{2018}+1}=1-\frac{1}{10^{2018}+1}\)
Có: \(B=\frac{10^{2018}}{10^{2019}+1}\)
\(\Rightarrow10B=\frac{10^{2019}}{10^{2019}+1}=1-\frac{1}{10^{2019}+1}\)
\(Vì10^{2018}+1< 10^{2019}+1nên\frac{1}{10^{2018}+1}>\frac{1}{10^{2019}+1}\)
\(\Rightarrow1-\frac{1}{10^{2018}+1}< 1-\frac{1}{10^{2019}+1}\)
\(hay10A< 10B\)
Suy ra: A < B
\(A=\frac{2018^{2019}+1}{2018^{2019}-2017}=\frac{2018^{2019}-2017+2018}{2018^{2019}-2017}=\frac{2018^{2019}-2017}{2018^{2019}-2017}+\frac{2018}{2018^{2019}-2017}=1+\frac{2018}{2018^{2019}-2017}\)\(B=\frac{2018^{2019}+2}{2018^{2019}-2016}=\frac{2018^{2019}-2016+2018}{2018^{2019}-2016}=\frac{2018^{2019}-2016}{2018^{2019}-2016}+\frac{2018}{2018^{2019}-2016}=1+\frac{2018}{2018^{2019}-2016}\)Ta có: \(2018^{2019}-2017< 2018^{2019}-2016\)
\(\Rightarrow\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)
\(\Rightarrow1+\frac{2018}{2018^{2019}-2017}>1+\frac{2018}{2018^{2019}-2016}\)
\(\Rightarrow A>B\)
Vậy...
Ta có :
\(A=\frac{2018^{2019}+1}{2018^{2019}-2017}=\frac{2018^{2019}-2017+2018}{2018^{2019}-2017}=1+\frac{2018}{2018^{2019}-2017}\)
\(B=\frac{2018^{2019}+2}{2018^{2019}-2016}=\frac{2018^{2019}-2016+2018}{2018^{2019}-2016}=1+\frac{2018}{2018^{2019}-2016}\)
Vì \(2018^{2019}-2017< 2018^{2019}-2016\)nên \(\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)hay \(A>B\)
~ Hok tốt ~