Tính các tổng sau
E= 1/15 +1/35 + ....+1/2499
*/* là phần nhaaz :v
Khog viết ra ps được . Sr :vv
3tick ?
K
Khách
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Những câu hỏi liên quan
3 tháng 4 2016
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
M=\(\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
3 tháng 4 2016
\(M=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\frac{50}{51}\)
\(M=\frac{25}{51}\)
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}.\frac{16}{51}\)
\(A=\frac{8}{51}\)
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(2A=\frac{1}{3}-\frac{1}{50}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{50}\right)\)
\(A=\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{50}\)
\(A=\frac{1}{6}-\frac{1}{100}=\frac{50}{300}-\frac{3}{300}=\frac{47}{300}\)
VT
15 tháng 4 2017
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
HỌC TỐT NHÉ BẠN!
29 tháng 5 2017
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}.\dfrac{16}{51}=\dfrac{8}{51}\)
Vậy \(C=\dfrac{8}{51}\)
QD
29 tháng 5 2017
\(\Rightarrow C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}.\dfrac{15}{51}\)
\(C=\dfrac{8}{51}\)
NV
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 5 2023
Olm sẽ hướng dẫn các em phương pháo giải tổng quát dạng này như sau:
Bước 1 phân tích số đã cho thành tích của các số nguyên tố
Bước 2 nhóm các thừa số nguyên tố thành 1 nhóm ta sẽ được tích của hai số cần tìm
2499 = 3 \(\times\) 7 \(\times\) 7 \(\times\) 17
2499 = ( 7 \(\times\) 7) \(\times\) ( 3 \(\times\) 17)
2499 = 49 \(\times\) 51
PV
3
30 tháng 4 2017
a) \(A=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(\Leftrightarrow A=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{102}{103}\)
\(\Leftrightarrow\) \(A=\dfrac{170}{103}\)
b) \(B=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}.\dfrac{16}{51}\)
\(B=\dfrac{8}{51}\)
2 tháng 6 2017
A = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
A = \(\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
A = \(\dfrac{5}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{100}+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{100}-\dfrac{1}{100}\right)-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-0-0-...-0-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{103}{103}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\dfrac{102}{103}\)
A = \(\dfrac{170}{103}\)
B = \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
B = \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
B = \(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
B = \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-0-0-...-0-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{17}{51}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\dfrac{16}{51}\)
B = \(\dfrac{8}{51}\)
NV
7
S
4 tháng 5 2023
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(=\dfrac{1}{3}-\dfrac{1}{51}\)
\(=\dfrac{16}{51}\)
24 tháng 5 2022
Tham khảo
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
MA
24 tháng 5 2022
Tham khảo
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
\(E=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)