\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
\(\frac{ }{\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}}\)
\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{8}{20\cdot28}\)
Bài 1 : Tính Nhanh
\(\frac{2}{5\cdot7}+\frac{5}{7\cdot12}+\frac{8}{12\cdot20}+\frac{3}{140}\)
Tính bằng phương pháp hợp lí
a> C = \(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{73\cdot76}\) b> D = \(\frac{5}{10\cdot11}+\frac{5}{11\cdot12}+\frac{5}{12\cdot13}+...+\frac{5}{99\cdot100}\)
NHANH LÊN MÌNH CẦN GẤP
\(20\%x+\frac{2}{5}x=x-4\)
\(\frac{\frac{2}{3}+\frac{2}{7}\cdot\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
giải hội mik mới cách làm lun nha
thực hiện phếp tính
\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot3+9\cdot12+12\cdot16+15\cdot20}\)
Rút gọn biểu thức \(Q=\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}+32\frac{15}{22}\right)}{\frac{7}{3\cdot13}+\frac{7}{13\cdot23}+\frac{7}{23\cdot33}}\) rồi tính 20% của Q ta đc bao nhiêu?
\(\frac{2}{4\cdot7}-\frac{2}{5\cdot9}+\frac{2}{7\cdot10}-\frac{2}{9\cdot13}+\frac{2}{10\cdot13}-\frac{2}{13\cdot17}+...+\frac{2}{301\cdot304}-\frac{2}{401\cdot405}CM:tich,tren,< \frac{1}{15}\)
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot30}+...+\frac{1}{73\cdot75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...3+\frac{4}{2008\cdot2010}\)