C = 1/10+1/15+1/21+1/28
K
Khách
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TP
1
27 tháng 8 2023
\(C\text{=}\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}\)
\(C\text{=}\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{10}+\dfrac{1}{15}\right)+\left(\dfrac{1}{21}+\dfrac{1}{28}\right)\)
\(C\text{=}\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}\)
\(C\text{=}\dfrac{3}{4}\)
29 tháng 3 2024
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
7 tháng 8 2024
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 = - \(\dfrac{54}{7}\)
N
2
Q
1
TN
10 tháng 5 2016
\(S=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2\times\frac{3}{16}\)
\(=\frac{3}{8}\)
DH
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\)
A = 2\(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\)+ \(\dfrac{1}{72}\))
A =2\(\times\)( \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\))
A = 2 \(\times\) ( \(\dfrac{1}{1}\)- \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))
A = 2\(\times\)( 1 - \(\dfrac{1}{9}\))
A = 2 \(\times\) \(\dfrac{8}{9}\)
A = \(\dfrac{16}{9}\)
ND
1
23 tháng 5 2021
B =1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
B = 1 - 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 - 1/28
B = 1 - ( 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 ) - 1/28
B = 1 - 1/28
B = 27/28
~ Hok T ~
PT
1
AH
Akai Haruma
Giáo viên
9 tháng 11 2021
Lời giải:
$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$
$=1-\frac{1}{9}=\frac{8}{9}$
$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$
HM
6
12 tháng 9 2015
Coi \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right).\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
HL
0
C = \(\frac{1}{10}+\frac{1}{15}\)\(+\frac{1}{21}+\frac{1}{28}\)
C = \(\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}\)
C = \(\frac{5-2}{2.5}+\frac{5-3}{3.5}+\frac{7-3}{3.7}+\frac{7-4}{4.7}\)
C = \(\frac{5}{2.5}-\frac{2}{2.5}+\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{3.7}-\frac{3}{3.7}+\frac{7}{4.7}-\frac{4}{4.7}\)
C = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{4}\)
C=\(\frac{1}{2}-\frac{1}{4}\)
C=\(\frac{1}{4}\)