A=-3+\(\dfrac{1}{1+1\dfrac{ }{4\dfrac{ }{3}}}=-3+\dfrac{1}{1+1\dfrac{ }{12}}=-3+\dfrac{1}{13\dfrac{ }{ }12}=-3+\dfrac{12}{13}=\dfrac{-27}{13}\)
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
8) \(A=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
9) \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2014}}+\dfrac{1}{3^{2015}}\)
10) \(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+...+\dfrac{1}{2004}}\)
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 ^^
Tìm x
a) \(\dfrac{3x-5}{\dfrac{2}{\dfrac{3}{4}}}=\dfrac{4}{5}\)
b) \(\dfrac{1-\dfrac{1}{3}+\dfrac{x}{1-\dfrac{1}{3}}}{1+\dfrac{1}{3}-\dfrac{1}{1+\dfrac{1}{3}}}\)
Bài 1: Thực hiện phép tính:
a) A = [6.(\(\dfrac{1}{3}\))3- 3 (-\(\dfrac{1}{3}\))+ 1 ] : (-\(\dfrac{1}{3}\)-1)
b) B = \(\dfrac{\dfrac{1}{39}-\dfrac{1}{6}-\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}:5\dfrac{1}{6}\)
\(tính:\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)
Tính nhanh A=\(\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)
B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
Tính \(\dfrac{B}{A}\)
Tính
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)
