Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Lời giải:
Ta có:
\(2018^{2018}(2019^{2019}+2019)=2018^{2018}.2019^{2019}+2018^{2018}.2019<2018^{2018}.2019^{2019}+2019^{2018}.2019 \)
\(< 2018^{2018}.2019^{2019}+2019^{2019}.2018\)
\(\Leftrightarrow 2018^{2018}(2019^{2019}+2019)< 2019^{2019}(2018^{2018}+2018)\)
\(\Rightarrow \frac{2018^{2018}}{2019^{2019}}< \frac{2018^{2018}+2018}{2019^{2019}+2019}\)
so sánh a và b biết a=2016/2017+2017/2018+2018/2019+2019/2016 và b=1/8+1/9+1/10+...+1/63
So sánh : \(\left(26^{2018}+3^{2018}\right)^{2019}\) và \(\left(26^{2019}+3^{2019}\right)^{2018}\)
So sánh:
A= (25^2018 + 9^2018)^2019
B= (25^2019 + 9^2019)^2018
lm cụ thể giúp mk vs nha. mk tik cho
x^2018+y^2018=2019^2018
x^2018+y^2018=2019^2018
(x+2018)^2020 - (x+2018)^2019 = 0
1.Cho A=20172018+20182019+20192020+20202021+2018
a)CMR: A chia hết cho 10
b)CMR 0,7 . A chia hết cho 7
(x+2018)^2020 - (x+2018)^2019 = 0
Cho a2018+b2018=a2019+b2019=1. Tính S=a2016+b2016
