\(\dfrac{2017}{2018}+\dfrac{2018}{2107}=1\dfrac{2017}{2018}+\dfrac{1}{2017}\)>\(1\dfrac{2017}{2018}+\dfrac{1}{2018}=2\)
=>\(1\dfrac{2017}{2018}+\dfrac{1}{2017}>2\)
=>\(\dfrac{2017}{2018}+\dfrac{2018}{2017}>2\)
\(\dfrac{2017}{2018}+\dfrac{2018}{2107}=1\dfrac{2017}{2018}+\dfrac{1}{2017}\)>\(1\dfrac{2017}{2018}+\dfrac{1}{2018}=2\)
=>\(1\dfrac{2017}{2018}+\dfrac{1}{2017}>2\)
=>\(\dfrac{2017}{2018}+\dfrac{2018}{2017}>2\)
So sánh A và B , biết
\(A=\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}\)
\(B=\dfrac{2017+2018+2019}{2018+2019+2020}\)
1. So sánh
A = \(\dfrac{2015}{2018^3}\) + \(\dfrac{2017}{2018^4}\) và B = \(\dfrac{2017}{2018^3}\) + \(\dfrac{2015}{2018^4}\)
Cho \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2017}{2^{2016}}+\dfrac{2018}{2^{2017}}\) . So sánh T và 3
so sánh 2017/2018 +2018/2017 với 2
\(\dfrac{X-1}{2019}+\dfrac{X-2}{2018}=\dfrac{X-3}{2017}+\dfrac{X-4}{2016}\)
B = \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^{2017}}+\dfrac{1}{2^{2018}}\)
tính B
Cho A=1/2+1/3+1/4+...+1/2019 và B=2018/1+2017/2+...+2/2017+1/2018. So sánh A/B với 1/2018
thực hiện phép tính.
a) b) C= 3+31+32+33+...+3100 c)\(\dfrac{2018.2019-1}{2018^2+2017}\)
\(\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
so sanh
A=2016/2017+2017/2018 va B=2016+2017/2017+2018