Vì \(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< 1\)
Ta có :
\(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< \dfrac{2017^{2018}-2+2019}{2017^{2019}-2+2019}=\dfrac{2017^{2018}+2017}{2017^{2019}+2017}=\dfrac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2018}+1\right)}=\dfrac{2017^{2017}+1}{2017^{2018}+1}=A\)
Vậy B < A
BT1: So sánh:
2) \(\dfrac{2017}{2018}+\dfrac{2018}{2019}\) VỚI \(\dfrac{2017+2018}{2018+2019}\)
BT1: Cho A = \(\dfrac{1}{2017}+\dfrac{2}{2017^2}+\dfrac{3}{2017^3}+...+\dfrac{2017}{2017^{2017}}+\dfrac{2018}{2017^{2018}}\)
Chứng minh rằng : A < \(\dfrac{2017}{2016^2}\)
a) Tính S = \(\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
b) Cho A = \(\dfrac{1}{2017}\)+ \(\dfrac{2}{2017^2}\) + \(\dfrac{3}{2017^3}\) + ... + \(\dfrac{2017}{2017^{2017}}\) + \(\dfrac{2018}{2017^{2018}}\)
Chứng minh tằng A < \(\dfrac{2017}{2016^2}\)
Nhanh lên nha chiều mình học rồi 
BT1: So sánh:
1) \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}\) VỚI 3
Không dùng máy tính cầm tay hãy so sánh
A = \(\dfrac{2017^{2017}+1}{2017^{2018}+1}\) và \(\dfrac{2017^{2016}+1}{2017^{2017}+1}\)
A= \(\dfrac{2016}{2017}+\dfrac{2017}{2018}\) và B = 2
Cho \(M=\dfrac{2018^{2017}+1}{2018^{2018}+1}\) và \(N=\dfrac{2018^{2016}+1}{2018^{2017}+1}\)
So sánh M và N
Giúp mk nha now!
Tính A+B
A= 1.99+2.98+3.97+.............+99.1
B=1.101+2.102+3.103+............+99.199
so sánh P và Q
P=2016/2017+2017/2018
Q= 2016+2017+2018/2017+2018+2019
\(Tính: B = \dfrac{2 - \dfrac{2}{19} + \dfrac{2}{43} - \dfrac{2}{2017}}{3 - \dfrac{3}{19} + \dfrac{3}{43} - \dfrac{3}{2017}} :\dfrac{4 - \dfrac{4}{29} + \dfrac{4}{41} - \dfrac{4}{2018}}{5 - \dfrac{5}{29} + \dfrac{5}{41} - \dfrac{5}{2018}} \)
