TY
5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
YB
1
LP
15 tháng 4 2015
C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
HT
1
4 tháng 5 2015
C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
DT
4
NH
27 tháng 7 2023
`A=7/(10*11)+7/(11*12)+7/(12*13)+...+7/(69*70)`
`1/7A=1/(10*11)+1/(11*12)+1/(12*13)+..+1/(69*70)`
`1/7A=1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70`
`1/7A=1/10-1/70`
`1/7A=7/70-1/70`
`1/7A=6/70`
`A=3/5`
27 tháng 7 2023
\(A=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)
\(A=7.\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{69.70}\right)\)
\(A=7\left(\dfrac{11-10}{10.11}+\dfrac{12-11}{11.12}+\dfrac{13-12}{12.13}+...+\dfrac{70-69}{69.70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(A=7.\dfrac{3}{35}=\dfrac{3}{5}\)
NC
4
6 tháng 5 2018
a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(3F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{1}{3}-\frac{1}{3}.\frac{1}{33}=\frac{1}{9}-\frac{1}{99}=\frac{11}{99}-\frac{1}{99}=\frac{10}{99}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}\)
\(A=\frac{7.6}{70}=\frac{1.6}{10}=\frac{1.3}{5}=\frac{3}{5}\)
6 tháng 5 2018
a, \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}-\frac{10}{33}\)
\(F=\frac{10}{99}\)
PT
0
31 tháng 1 2022
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
PN
2
11 tháng 1 2018
A = [ 1 + (-3) ] + [ 5 + (-7) ]+....+ [999 + (-1001) ] (Có 251 cặp )
A = (-2) + (-2) +.....+ (-2) (Có 251 số -2)
A = (-2).251
A = -502
Vậy A = -502
18 tháng 7 2015
số số hạng là: (100-1):1+1=100 (số)
tổng: (100+1) x 100 : 2=5050
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}=\frac{3}{5}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=1\left(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\right)\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)