Question

so sánh A và B

^{a, A= 102017 +1/ 102018+1 , B =}10²⁰¹⁸ +1/ 10²⁰¹⁹+1

Answer 1

\(A=\dfrac{10^{2017}+1}{10^{2018}+1}\)

=>\(10A=\dfrac{10^{2018}+1+9}{10^{2018}+1}=1+\dfrac{9}{10^{2018}+1}\)

\(B=\dfrac{10^{2018}+1}{10^{2019}+1}\)

=>\(10B=\dfrac{10^{2019}+1+9}{10^{2019}+1}=1+\dfrac{9}{10^{2019}+1}\)

Do đó:\(10B< 10A\)=>\(B< A\)

Answer 2

\(A=\dfrac{10^{2017}+1}{10^{2018}+1}\)

\(10A=\dfrac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\dfrac{10^{2018}+10}{10^{2018}+1}=\dfrac{10^{2018}+1+9}{10^{2018}+1}=\dfrac{10^{2018}+1}{10^{2018}+1}+\dfrac{9}{10^{2018}+1}=1+\dfrac{9}{10^{2018}+1}\)\(B=\dfrac{10^{2018}+1}{10^{2019}+1}\)

\(10B=\dfrac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\dfrac{10^{2019}+10}{10^{2019}+1}=\dfrac{10^{2019}+1+9}{10^{2019}+1}=\dfrac{10^{2019}+1}{10^{2019}+1}+\dfrac{9}{10^{2019}+1}=1+\dfrac{9}{10^{2019}+1}\)Vì \(1+\dfrac{9}{10^{2018}+1}>1+\dfrac{9}{10^{2019}+1}\)

Nên \(10A>10B\)

Nên \(A>B\)