1/7*12+1/12*17+...+1/2017*2012
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J
1+2-3-4-5+6+7-8-9-10+11+12-13-14-15+...+2011+2012-2013-2014-2015+2016+2017-2018-2019-2020 giup mik v
1
AH
Akai Haruma
Giáo viên
31 tháng 12 2023
Lời giải:
$A=(1+2-3-4-5)+(6+7-8-9-10)+(11+12-13-14-15)+....+(2011+2012-2013-2014-2015)+(2016+2017-2018-2019-2020)$
$=(-9)+(-14)+(-19)+....+(-2019)+(-2024)$
$=-(9+14+19+...+2019+2024)$
Số số hạng: $(2024-9):5+1=404$
$A=-(2024+9).404:2=-410666$
C
23 tháng 9 2018
\(\dfrac{12}{17}.\dfrac{5}{7}-\left(-\dfrac{12}{17}\right).\dfrac{1}{7}+\dfrac{1}{17+7}\)
\(=\dfrac{12}{17}.\dfrac{5}{7}+\dfrac{12}{17}.\dfrac{1}{7}+\dfrac{1}{24}\)
\(=\dfrac{12}{17}\left(\dfrac{5}{7}+\dfrac{1}{7}\right)+\dfrac{1}{24}\)
\(=\dfrac{12}{17}.\dfrac{6}{7}+\dfrac{1}{24}\)
\(=\dfrac{72}{119}+\dfrac{1}{24}=\dfrac{1847}{2856}\)
NN
0
12 tháng 3 2020
Ta có:
\(B=\frac{10}{7.12}+\frac{10}{12.17}+...+\frac{10}{2017.2022}\)
\(\Rightarrow B=2.\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{2017.2022}\right)\)
\(\Rightarrow B=2.\left(\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{2022-2017}{2017.2022}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2017}-\frac{1}{2022}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{7}-\frac{1}{2022}\right)\)
\(\Rightarrow B=\frac{2015}{7077}\)
TD
4
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
N
4
14 tháng 4 2016
2/5 + 3/4 = 23/40
7/12 - 2/7 + 1/12 = 25/84 + 1/12 = 8/21
12/17 - 5/17 - 4/17 = 3/17
OK
14 tháng 4 2016
\(\frac{2}{5}+\frac{3}{4}=\frac{8}{20}+\frac{15}{20}=\frac{23}{20}\)
\(\frac{7}{12}-\frac{2}{7}+\frac{1}{12}=\left(\frac{7}{12}+\frac{1}{12}\right)-\frac{2}{7}=\frac{2}{3}-\frac{2}{7}=\frac{14}{21}-\frac{6}{21}=\frac{8}{21}\)
\(\frac{12}{17}-\frac{5}{17}-\frac{4}{17}=\frac{12-5-4}{17}=\frac{3}{17}\)
\(\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+...+\frac{1}{2017\cdot2022}\)
\(=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{2017\cdot2022}\right)\)
\(=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{2}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2017}-\frac{1}{2022}\right)\)
\(=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{2022}\right)\)
\(=\frac{1}{4}\cdot\frac{2015}{14154}\)
\(=\frac{2015}{56616}\)