\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)
Tự so sánh
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)
Từ (1) và (2) suy ra 100A > 100B hay A > B
\(B=\frac{100^{2018}+1}{100^{2019}+1}< \frac{100^{2018}+1+99}{100^{2019}+1+99}=\frac{100^{2018}+100}{100^{2019}+100}=\frac{100\left(100^{2017}+1\right)}{100\left(100^{2018}+1\right)}=\frac{100^{2017}+1}{100^{2018}+1}=A\)
\(\Rightarrow B< A\)
1-A=1--\(\frac{100^{2017}+1}{100^{2018}+1}=\frac{100^{2018}+1}{100^{2018}+1}-\frac{100^{2017}+1}{100^{2018}+1}=\frac{2}{100^{2018}+1}\)
1-B=1--\(\frac{100^{2018}+1}{100^{2019}+1}=\frac{100^{2019}+1}{100^{2019}+1}-\frac{100^{2018}+1}{100^{2019}+1}=\frac{2}{100^{2019}+1}\)
Do \(\frac{2}{100^{2018}+1}>\frac{2}{100^{2019}+1}\)
Suy Ra A<B
