Các câu hỏi tương tự
Cho A= 2018^100 + 2018^96 +...+2018^4+1/2018^102+2018^100+....=2018^2+1
CMR 4A < (0,1)^6
GIÚP MÌNH NHA, THANKS
Cho A= 2018^100 +2018^96+...+2018^4+1/2018^102+2018^100+...+2018^2+1
CMR 4A< (0,1)^6
\(Cho\)\(A=\)\(\frac{2018^{100}+2018^{96}+....+2018^4+1}{2018^{102}+2018^{100}+...+2018^2+1}\)
\(CMR\)\(:\)\(4A< \left(0,1\right)^6\)
GIÚP MÌNH NHA , MÌNH ĐANG CẦN GẤP ....
â , tính M = \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).......\left(1+\frac{1}{2017}\right)\left(1+\frac{1}{2018}\right)\)
b , Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
c , B = \(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2017}+\frac{1}{2018}.tinh\left(\frac{A}{B}\right)^{2018}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh: \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Cho \(A=1.2.3...2018\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\).CMR: \(A⋮2019\)
cho \(\frac{a}{b}=\frac{c}{d}\left(b,d\ne0;b\ne d,-d\right)\)
Chứng minh \(\left(\frac{a-b}{c-d}\right)^{2018}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)
Cho\(a+b+c=2018\) và\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}\)CMR: \(\orbr{\begin{cases}x=2018\\y=2018\end{cases}}\)hoặc z=2018
Tính :
a) \(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+...+\left(2014\times2015\right)^{-1}\).
b) \(\text{B}=\frac{2018+\frac{2017}{2}+\frac{2016}{3}+\frac{2015}{4}+...+\frac{2}{2017}+\frac{1}{2018}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2018}+\frac{1}{2019}}\).