Tính tổng sau:
S = \(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+...+\(\dfrac{1}{300}\)
K
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VT
25 tháng 6 2023
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
HB
2
18 tháng 4 2020
\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)
\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)
\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)
Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)
2 tháng 5 2023
a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\).
Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)
\(A=\dfrac{n+4}{n+1}\) làm tương tự.
b) Dễ thấy các số ở mẫu có thể viết dưới dạng:
\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)
\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)
\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)
...
\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)
Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
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 = - \(\dfrac{54}{7}\)
6 tháng 9 2017
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)
\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)\)
\(=2\cdot\dfrac{23}{50}=\dfrac{23}{25}\)
Vậy A = \(\dfrac{23}{25}\).
NN
6 tháng 9 2017
It's very easy :)))))
Call expression is A, we have:
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}.\)
\(\Rightarrow2A=2\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\right).\)
\(\Rightarrow2A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}.\)
\(\Rightarrow2A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}.\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\right).\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right).\)
\(\Rightarrow2A=2\left(\dfrac{25}{50}-\dfrac{2}{50}\right).\)
\(\Rightarrow2A=2.\dfrac{23}{50}=\dfrac{23}{25}.\)
Vậy.....
27 tháng 9 2021
Ta có:
\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)
\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)
\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)
\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)
\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)
\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)
\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)
\(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)
\(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)
\(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)
\(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)
\(=\dfrac{28}{36}=\dfrac{7}{9}\)
Vậy: \(A=\dfrac{7}{9}\)
17 tháng 7 2023
\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)
\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)
Vậy D=1/4
\(S=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)
\(S=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)
\(S=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)
\(S=2.(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25})\)
\(S=2.(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25})\)
\(S=2.(\dfrac{1}{4}-\dfrac{1}{25})\)
\(S=2.\dfrac{21}{100}\)
\(S=\dfrac{21}{50}\)
\(\Rightarrow S=\dfrac{21}{50}\)
S = \(\dfrac{1}{10}\) + \(\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)
S = \(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)
S = \(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)
S = 2 (\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\))
S = 2 (\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\))
S = 2 . (\(\dfrac{1}{4}-\dfrac{1}{25}\))
S = 2 . \(\dfrac{21}{100}\)
=> S = \(\dfrac{21}{50}\)