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

Làm sao để tách 1/2499 v

Toán lớp 1?

=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!

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

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

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

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

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

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

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

Bài giải:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

C=1/15+1/35+1/63+..+1/2499

   =1/3.5+1/5.7+1/7.9+...+1/49.51

  =1/2(2/3.5+2/5.7+2/7.9+...+2/49.51)

  =1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/49-1/51)

  =  1/2.(1/3-1/51)

  =1/2.16/51 

  =8/51

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

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

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

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

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

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

bạn tách cái số 2499 kiểu j thế

1. máy tính

2.nhẩm

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

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

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

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

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

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

Vậy \(C=\dfrac{8}{51}\)

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

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

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

\(C=\dfrac{1}{2}.\dfrac{15}{51}\)

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

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

\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{49\cdot51}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)

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

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

\(=\dfrac{1}{2}\left(\dfrac{17}{51}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{16}{51}\)

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

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)

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

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

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

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)

C=\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+......+\(\frac{1}{49.51}\)

C=(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+.......+\(\frac{2}{49.51}\)) :2

C=(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+....+\(\frac{1}{49}\)-\(\frac{1}{51}\)) :2

C=(\(\frac{1}{3}\)-\(\frac{1}{51}\)) :2

C=\(\frac{16}{51}\):2=\(\frac{8}{51}\)