Cho A=1\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)......\(+\)\(\frac{1}{2013}\)\(-\)\(\frac{1}{2014}\)

Cho B=\(\frac{1}{2016}+\frac{1}{2018}+\frac{1}{2020}+.....+\frac{1}{4028}\)

Tính A \(\div\)B

1.

\(\frac{1}{2}\)+\(\frac{2}{3}\)+\(\frac{3}{4}\)

\(\frac{4}{5}\)+\(\frac{4}{3}\)+\(\frac{6}{7}\)

\(\frac{1}{4}\)+\(\frac{2}{5}\)+\(\frac{1}{3}\)

\(\frac{3}{5}\)+\(\frac{5}{6}\)+\(\frac{7}{9}\)

\(\frac{8}{9}\)-\(\frac{5}{6}\)

\(\frac{99}{108}\)-\(\frac{40}{72}\)

\(\frac{16}{17}\)-\(\frac{12}{13}\)

\(\frac{99}{153}\)-\(\frac{75}{125}\)

Đề test thành viên : Team Math is easy 

Bài 1: Tính nhanh:

A. \(\frac{1}{1\times2}\)\(+\frac{1}{2\times3}\)\(+\frac{1}{3\times4}\)\(+............+\frac{1}{1999\times2000}\)

B. \(\frac{1}{2}\)\(+\frac{1}{6}\)\(+\frac{1}{12}\)\(+\frac{1}{20}\)\(+...........+\frac{1}{39800}\)

C. \(\frac{1}{2}\)\(+\frac{1}{4}\)\(+\frac{1}{8}\)\(+\frac{1}{16}\)\(+\frac{1}{32}\)\(+\frac{1}{64}+\frac{1}{128}\)\(+\frac{1}{256}\)\(+\frac{1}{512}\)

# Phó team

\(\frac{10}{18}\)+\(\frac{4}{9}\)+\(\frac{26}{10}\)+\(\frac{12}{5}\)+\(\frac{9}{15}\)

\(\frac{10}{18}\)+\(\frac{4}{9}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+....+\(\frac{1}{64}\)+\(\frac{1}{128}\)

\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+.....+\(\frac{2}{90}\)+\(\frac{2}{100}\)

bài này tính nhanh nha

\(8\frac{7}{10}\)+\(2\frac{3}{4}\)=

\(2\frac{7}{8}\)-\(1\frac{3}{8}\)=

\(3\frac{1}{3}\)x\(2\frac{4}{5}\)=

\(4\frac{1}{2}\):\(2\frac{1}{7}\)=

\(2\frac{1}{5}\)+\(1\frac{7}{10}\)=

\(8\frac{1}{4}\)-\(5\frac{1}{2}\)=

\(3\frac{4}{5}\)x\(\frac{10}{21}\)=

\(1\frac{4}{7}\):\(2\frac{3}{8}\)

viết 5 số tiếp theo của dãy

0;2;4;6;12;22.......

\(\frac{1}{1}\);\(\frac{2}{1}\);\(\frac{1}{2}\);\(\frac{3}{1}\);\(\frac{2}{2}\);\(\frac{1}{3}\);\(\frac{4}{1}\);\(\frac{3}{2}\);\(\frac{2}{3}\);\(\frac{1}{4}\).........

tính giá trị biểu thức:

B=(2017-\(\frac{1}{4}\)-\(\frac{2}{5}\)-\(\frac{3}{6}\)-\(\frac{4}{7}\)-.....-\(\frac{2017}{2020}\)) : (\(\frac{1}{20}\)+\(\frac{1}{25}\)+\(\frac{1}{30}\)+\(\frac{1}{35}\)+.....+\(\frac{1}{10100}\))