Question

[trần võ thị mỹ lên]

SO SÁNH A VÀ B, BIẾT: A=2018²⁰¹⁹+1/2O18²⁰²⁰+1

                                                      B=2018²⁰¹⁸+1/2018²⁰¹⁹+1

Answer 1

[Linh Hương]

So sánh :

A = \(\frac{2018^{2019}+1}{2018^{2020}+1}\)và B = \(\frac{2018^{2018}+1}{2018^{2019}+1}\)

Ta có : 

+ , A = \(\frac{2018^{2019}+1}{2018^{2020}+1}\)

2018A =  \(\frac{2018\left(2018^{2019}+1\right)}{2018^{2020}+1}\)

2018A = \(\frac{2018^{2020}+2018}{2018^{2020}+1}\)

2018A = \(1\frac{2018}{2018^{2020}+1}\)

+ , B = \(\frac{2018^{2018}+1}{2018^{2019}+1}\)

2018B = \(\frac{2018\left(2018^{2018}+1\right)}{2018^{2019}+1}\)

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

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

Ta thấy : 

2018²⁰¹⁹ + 1 < 2018²⁰²⁰ + 1

= > \(\frac{2018}{2018^{2019}+1}\)> \(\frac{2018}{2018^{2020}+1}\)

= > B > A 

Vậy B > A