\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(A=\frac{1}{2}.\frac{16}{51}\)

\(A=\frac{8}{51}\)

\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)

\(2A=\frac{1}{3}-\frac{1}{50}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{50}\right)\)

\(A=\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{50}\)

\(A=\frac{1}{6}-\frac{1}{100}=\frac{50}{300}-\frac{3}{300}=\frac{47}{300}\)

Bài của Nguyễn Thanh Hiền đúng rồi. Bài của mk sai nha

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 

\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(=\dfrac{1}{3}-\dfrac{1}{51}\)
\(=\dfrac{16}{51}\)

Làm sao để tách 1/2499 v

Toán lớp 1?

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

= \(\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

Bài giải:

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{8}{51}\)

=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!

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{49.51}\right)\)

=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

=\(\frac{1}{2}.\frac{16}{51}\)

=\(\frac{8}{51}\)

bạn tách cái số 2499 kiểu j thế

1. máy tính

2.nhẩm

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) \(B=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)

\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)

\(B=\dfrac{1}{2}.\dfrac{16}{51}\)

\(B=\dfrac{8}{51}\)

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 = \(\dfrac{8}{51}\)

a) 1/n-1/n+a=a/n.(n+a)

=(n+a)-n/n.(n+a)=n-n+a/n.(n+a)

=a/n.(n+a)

b)1/15+1/35+...+1/2499

=1/3.5+1/5.7+......+1/49.51

=1.2/2.3.5+1.2/2.5.7+.....+1.2/2.49.51

=1/2(2/3.5+2/5.7+.....+2/49.51)

=1/2(1/3-1/5+1/5-1/7+.......+1/49-1/50)

=1/2(1/3-1/50)

=1/2(50/150-3/150)

=1/2.47/150

=47/300

Ta có : 

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\)\(\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{2499}\right)\) ( bước này hơi khó hiểu tí nhé ) 

\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\) ( phân tích mẫu ) 

\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\) ( áp dụng công thức thoi ) 

\(=\)\(\frac{1}{2}\left(1-\frac{1}{51}\right)\) ( loại bỏ nhưng phân số đối nhau ) 

\(=\)\(\frac{1}{2}.\frac{50}{51}\)

\(=\)\(\frac{25}{51}\)

Chúc bạn học tốt ~ 

ĐẶT \(A\)\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)

\(2.A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)

\(2.A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(2.A=1-\frac{1}{51}\)

\(2.A=\frac{50}{51}\)

\(\Rightarrow A=\frac{50}{51}\div2=\frac{25}{51}\)

Cảm ơn Nhìu nha Phùng Minh Quân <3