Tính \(A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{15}+...+\dfrac{1}{95}.\dfrac{1}{99}\)
a) \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}\)\(-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
b)\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+......+\dfrac{1}{255.257}\)
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
b, \(\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+......+\dfrac{1}{3.2}+\dfrac{1}{2.1}\)
\(=\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+.........+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\))
\(=\dfrac{1}{99}+\dfrac{1}{99}-1=\dfrac{2}{99}-1=\dfrac{-97}{99}\)
Chúc bạn học tốt!!!
c/ \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1}{155\cdot257}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}+\dfrac{2}{255\cdot257}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}+...+\dfrac{1}{255}-\dfrac{1}{257}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{257}\right)=\dfrac{1}{2}\cdot\dfrac{256}{257}=\dfrac{128}{257}\)
p/s: thấy bn kia hơn 20' r` k lm tiếp nên t lm lun nhé!
2 Chứng minh:
a) \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{4}.(\dfrac{1}{n}-\dfrac{1}{n+4})\) b)Tính A=\(\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}\)
a, Ta có: \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{n.\left(n+4\right)}=\dfrac{1}{4}.\dfrac{4}{n.\left(n+1\right)}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
Vậy \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
b, \(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}=4.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4.\dfrac{32}{99}=\dfrac{128}{99}\)
Vậy \(A=\dfrac{128}{99}\)
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
tính nhanh.
0.125*\(\dfrac{3}{7}\)-\(\dfrac{1}{8}\)*\(\dfrac{11}{7}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(0,125.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}\left(\dfrac{3}{7}-\dfrac{11}{7}\right)=\dfrac{1}{8}.-\dfrac{8}{7}=-\dfrac{1}{7}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
tinh nhanh
A=\(\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
B=\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}\)
giup minh nhe minh dang can gap
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
tinh nhanh
A=\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}
B=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}
giup minh nhe minh dang can gap
Cho A=\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{11}+\dfrac{1}{15}+\dfrac{1}{19}+...+\dfrac{1}{99}\)
Tính A.
Giúp mình với:
a) A=\(\dfrac{1}{3}\)-\(\dfrac{3}{5}\)+\(\dfrac{5}{7}\)-\(\dfrac{7}{9}\)+\(\dfrac{9}{11}\)-\(\dfrac{11}{13}\)+\(\dfrac{13}{15}\)+\(\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
b) B= \(\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-...-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)
a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
A=0+0+0+...+0+\(\dfrac{13}{15}\)
A=\(\dfrac{13}{15}\)
b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)
Thực hiện phép tính:
a) \(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(\dfrac{-4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
b)\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)