Violympic toán 7
Các câu hỏi tương tự
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
Tính
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)
tìm x biết (\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2018}\)).x=\(\dfrac{2017}{1}+\dfrac{2016}{2}+\dfrac{2015}{3}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
Tìm x :
a) \(3-\dfrac{2}{2x-3}=\dfrac{2}{5}+\dfrac{1}{2x-3}-\dfrac{3}{2}\)
b) \(7.\left(x.1\right)+2x.\left(1-x\right)=0\)
c) \(\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}=\dfrac{x+10}{2009}+\dfrac{x+11}{2008}+\dfrac{x+12}{2007}\)
Có 3 ý thui các bạn hãy giúp mk!!!!!!
CM:
\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{4}\)
Tính: \(D=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{\dfrac{2011}{1}+\dfrac{2010}{2}+\dfrac{2009}{3}+...+\dfrac{1}{2011}}\)
CMR: \(C=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)
So sánh:
a) A dfrac{1}{2^1}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{49}}+dfrac{1}{2^{50}} với 1
b) B dfrac{1}{3^1}+dfrac{1}{3^2}+dfrac{1}{3^3}+...+dfrac{1}{3^{99}}+dfrac{1}{3^{100}} với dfrac{1}{2}
c) C dfrac{1}{4^1}+dfrac{1}{4^2}+dfrac{1}{4^3}+...+dfrac{1}{4^{999}}+dfrac{1}{4^{1000}} với dfrac{1}{3}
Cần gấp ạ ^^ Cảm ơn trước ^^
Đọc tiếp
So sánh:
a) A = \(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\) với 1
b) B = \(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\) với \(\dfrac{1}{2}\)
c) C = \(\dfrac{1}{4^1}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{999}}+\dfrac{1}{4^{1000}}\) với \(\dfrac{1}{3}\)
Cần gấp ạ ^^ Cảm ơn trước ^^