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7 tháng 9 2016
a) i don't know
b) \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{28.29}\)
\(5.\left(1-\frac{1}{2}\right)+5.\left(\frac{1}{2}-\frac{1}{3}\right)+5.\left(\frac{1}{3}-\frac{1}{4}\right)+...+5.\left(\frac{1}{28}-\frac{1}{29}\right)\)
\(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{28}-\frac{1}{29}\right)\)
\(5.\left(1-\frac{1}{29}\right)\)
\(5.\frac{28}{29}\)
\(\frac{140}{29}\)
7 tháng 9 2016
c ) \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{28.30}\)
\(2.\left(\frac{1}{2}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{6}\right)+...+2.\left(\frac{1}{28}-\frac{1}{30}\right)\)
\(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(2.\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(2.\frac{7}{15}=\frac{14}{15}\)
d) đề là thế này nè : 4/6 + 4/12 + 4/20 + 4/30 + 4/42 + 4/56
4/2.3 + 4/3.4 + 4/4.5 + 4/5.6 + 4/6.7 + 4/7.8
4.(1/2-1/3) + 4.(1/3-1/4 ) + 4.(1/4-1/5) + 4.(1/5-1/6 ) + 4.(1/6 - 1/7 ) + 4.(1/7-1/8 )
4.(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8)
4.(1/2-1/8)
4.1/16
1/4
AM
14 tháng 6 2015
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
NT
0
TA
9 tháng 6 2023
a, \(2\dfrac{1}{3}+4\dfrac{1}{5}+4\dfrac{1}{3}\)
\(=\dfrac{7}{3}+\dfrac{21}{5}+\dfrac{13}{3}\)
\(=\dfrac{1}{3}\left(7+13\right)+\dfrac{21}{5}\)
\(=\dfrac{20}{3}+\dfrac{21}{5}=\dfrac{100+63}{15}=\dfrac{163}{15}\)
b, \(5\dfrac{3}{4}-4\dfrac{1}{2}.3\dfrac{7}{8}\)
\(=\dfrac{23}{4}-\dfrac{9}{2}.\dfrac{31}{8}\)
\(=\dfrac{23}{4}-\dfrac{279}{16}=\dfrac{92-279}{16}=-\dfrac{187}{16}\)
c, \(1\dfrac{1}{2}.3\dfrac{2}{3}.4\dfrac{3}{4}\)
\(=\dfrac{3}{2}.\dfrac{11}{3}.\dfrac{19}{4}=\dfrac{209}{8}\)
d, \(6\dfrac{4}{5}:2\dfrac{3}{4}:1\dfrac{1}{2}\)
\(=\dfrac{34}{5}:\dfrac{11}{4}:\dfrac{3}{2}\)
\(=\left(\dfrac{34}{5}.\dfrac{4}{11}\right).\dfrac{2}{3}=\dfrac{136}{55}.\dfrac{2}{3}=\dfrac{272}{165}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 10 2023
[4\(\dfrac{1}{5}\) - 2\(\dfrac{2}{5}\)] x 8\(\dfrac{5}{6}\)
= [\(\dfrac{21}{5}\) - \(\dfrac{12}{5}\)] x \(\dfrac{53}{6}\)
= \(\dfrac{9}{5}\) x \(\dfrac{53}{6}\)
= \(\dfrac{159}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 10 2023
[5\(\dfrac{1}{3}\) - 2\(\dfrac{2}{3}\)]: \(\dfrac{3}{7}\)
= [\(\dfrac{16}{3}\) - \(\dfrac{8}{3}\)]: \(\dfrac{3}{7}\)
= \(\dfrac{8}{3}\) : \(\dfrac{3}{7}\)
= \(\dfrac{56}{9}\)
LB
11 tháng 5 2017
a)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(\frac{2.3\left(1.2\right)+2.3\left(2.4\right)+2.3\left(3.6\right)+2.3\left(4.8\right)+2.3\left(5.10\right)}{3.4\left(3.4+6.8+9.12+12.16+15.20\right)}\)
\(=\frac{\left(3.4+6.8+9.12+12.16+15.20\right)}{2.3\left(3.4+6.8+9.12+12.16+15.20\right)}=\frac{1}{2.3}=\frac{1}{6}\)
DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
CT
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 10 2024
B = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\) (như này mới đúng em ơi)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 10 2024
C = 8 : \(\dfrac{4}{5}\) - 7\(\dfrac{1}{5}\) x \(\dfrac{5}{4}\)
C = 8 x \(\dfrac{5}{4}\) - \(\dfrac{36}{5}\) x \(\dfrac{5}{4}\)
C = \(\dfrac{5}{4}\) x (8 - \(\dfrac{36}{5}\))
C = \(\dfrac{5}{4}\) x (\(\dfrac{40}{5}\) - \(\dfrac{36}{5}\))
C = \(\dfrac{5}{4}\) x \(\dfrac{4}{5}\)
C = 1
\(\left(-2\right)+\left(-4\right)\)
\(=-\left(2+4\right)\)
\(=-6\)
= -6