Các câu hỏi tương tự
Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
1)\(A=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
tính B:A
2)\(A=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{4026}\)
\(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\)
So sánh \(\frac{A}{B}và1\frac{2013}{2014}\)
1. Tính nhanh
a, \(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\)
b, \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c,\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
a)Tính\(\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}-32\frac{15}{22}\right)}{\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}}\)
b) Chứng tò rằng:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Cho M=\(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\)
N=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Tính N:M
Cho \(M=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\) và \(N=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}\)
Tính N : M
tính
A=\(\frac{\frac{1}{199}+\frac{1}{198}+....+\frac{1}{1}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}}\)
cho A /\(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
cho B:\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
Tính A:B
1. Tính \(\frac{A}{B}\) biết:
a, A=\(\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
b, A=\(\frac{1}{1.400}+\frac{1}{2.401}+\frac{1}{3.402}+...+\frac{1}{101.500}\)
B=\(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{399.500}\)