=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

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?

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

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

M=\(\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)

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

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

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

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

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

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

M=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

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

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

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

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

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

( k mình nha)

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

a) A x 3/5 = 3/1.4 + 3/4.7 + 3/7.10 + ...+ 3/100.103

               = 1 - 1/4 + 1/4 - 1/7 + 1/7 -1/10 + ... + 1/100 - 1/103

               = 1 - 1/103 

               = 102/103   => A = 102/103 : 3/5 = 170/103

b) B = 1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/49.51

 B x2 = 2/3.5 + 2/5.7 + 2/7.9 + ...+ 2/49.51

         = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1.49 - 1/51

         = 1/3 - 1/51

        = 16/51

=> B = 16/51 : 2 = 8/51 

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

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

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

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

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

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

Bài giải:

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

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

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

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

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