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\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)
=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)
=> 411 < 10x < 521
=> x \(\in\){ 42,43,44,...,52}
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(1-\frac{1}{10}\right)=-\frac{9}{10}\)
\(N=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{10}{33}\)
\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4970}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{70.71}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{70}-\frac{1}{71}\)
\(M=1-\frac{1}{71}\)
\(M=\frac{70}{71}\)
\(N=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(N=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(N=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\frac{10}{33}\)
\(N=\frac{10}{99}\)
#)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+\(\frac{3}{5}\)+\(\frac{14}{15}\).
=\(\frac{8}{5}\)+\(\frac{14}{15}\).
=\(\frac{15}{38}\)
A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
=1/2.3+1/3.6+1/6.6+1/6.10+1/10.9+1/9.14
=1/2-1/3+1/3-1/6+1/6-1/6+1/6-1/10+1/10-1/9+1/9-1/14
=1/2-1/14
=6/14=3/7
\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot6}+\frac{1}{6\cdot6}+\frac{1}{6\cdot10}+\frac{1}{10\cdot9}+\frac{1}{9\cdot14}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)