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Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
=2/10+3/10+4/10+......+13/10
=\(\frac{2+3+4+......+13}{10}\)
=90/10=9
k cho mình nha
\(-0,8\cdot1\frac{9}{25}-\frac{4}{5}\cdot\frac{64}{25}\)
\(=\frac{-4}{5}\cdot\frac{34}{25}+\frac{-4}{5}\cdot\frac{64}{25}\)
\(=\frac{-4}{5}\left(\frac{34}{25}+\frac{64}{25}\right)\)
\(=\frac{-4}{5}\cdot\frac{98}{25}\)
\(=\frac{-392}{125}\)
Chắc sai =))
F=(3/1.8+3/8.15+...+3/106.113)-(25/50.55+25/55.60+...+25/95.100) (1)
Đặt:
A=(3/1.8+3/8.15+3/15.22+...+3/106.113)
=3/7.(1-1/8+1/8-1/15+...+1/106-1/113)
=3/7.(1-1/113)
=3/7.112/113
=336/791. (2)
B=25/50.55+25/55.60+...+25/95.100
=25/5.(1/50-1/55+1/55-1/60+...+1/95-1/100)
=5.(1/50-1/100)
=1/20 (3)
Thay (2),(3) vào (1) ta được:
F=336/791-1/20
=5929/6720.
#)Giải :
\(\frac{1}{15}+\frac{4}{30}+\frac{9}{45}+\frac{16}{60}+...+\frac{81}{135}=\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+...+\frac{9}{15}=\frac{45}{15}=3\)
Dễ ẹc ak :v rút gọn là ra
=(\(\frac{1}{15}\)+\(\frac{4}{30}\)+\(\frac{16}{60}\)+\(\frac{64}{120}\))+(\(\frac{9}{45}\)+\(\frac{36}{90}\))+(\(\frac{25}{75}\)+\(\frac{81}{135}\))
=(\(\frac{8}{120}\)+\(\frac{16}{120}\)+\(\frac{32}{120}\)+\(\frac{64}{120}\))+(\(\frac{18}{90}\)+\(\frac{36}{90}\))+\(\frac{14}{15}\).
=1+\(\frac{3}{5}\)+\(\frac{14}{15}\).
=\(\frac{8}{5}\)+\(\frac{14}{15}\).
=\(\frac{15}{38}\)
\(\frac{\frac{2}{33}+\frac{2}{25}+\frac{2}{19}}{\frac{3}{33}+\frac{3}{25}+\frac{3}{19}}=\frac{2\left(\frac{1}{33}+\frac{1}{25}+\frac{1}{19}\right)}{3\left(\frac{1}{33}+\frac{1}{25}+\frac{1}{19}\right)}=\frac{2}{3}\)
= \(\frac{2\left(\frac{1}{33}+\frac{1}{25}+\frac{1}{19}\right)}{3\left(\frac{1}{33}+\frac{1}{25}+\frac{1}{19}\right)}=\frac{2}{3}\)
\(\frac{25}{72}+\frac{25}{90}+\frac{25}{110}+...+\frac{25}{9900}\)
= \(25.\left(\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+...+\frac{1}{9900}\right)\)
= \(25.\left(\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{99.100}\right)\)
= \(25.\left(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(25.\left(\frac{1}{8}-\frac{1}{100}\right)\)
= \(25.\frac{23}{200}\)
= \(\frac{23}{8}\)