Ta có:

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)

Coi \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\)

\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}+\frac{2}{55.57}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(=\frac{1}{3}-\frac{1}{57}\)

\(=\frac{19}{57}-\frac{1}{57}=\frac{18}{570}=\frac{6}{19}\)

\(\Rightarrow A=\frac{6}{19}:2=\frac{3}{19}\)

Vậy tổng trên bằng \(\frac{3}{19}\)

1/15 +1/35 +1/63 + 1/99 +...+1/2915 +1/3135

=1/3x5+1/5x7+1/7x9+....+1/53x55+1/55x57

=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/53-1/55+1/55-1/57

=1/3-1/57

=6/19 nhé

Ta có:1/15+1/35+1/63+1/99+...+1/2915+1/3135=1/3*5+1/5*7+1/7*9+1/9*11+...+1/53*55+1/55*57

=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55+1/55-1/57

=1/3-1/57=19/57-1/57=18/57

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+..+\frac{1}{3135}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+..+\frac{1}{55.57}\)
\(\Rightarrow\)\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)\(+...+\frac{1}{55}-\frac{1}{57}\)\(=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)

 

Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135

         =1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57

         =1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)

         =1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

         =1/2(1/3-1/57)

         =1/2(19/57-1/57)

         =1/2.18/57

         =3/19

Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19

Mik viết thế này mong bạn thông cảm nha!!

chúc bạn hok tốt!!

Bạn nhớ k cho mik một cái đúng nha!!

Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)

\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)

\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)

1/15 + 1/35 + 1/63 + 1/99 + ...+ 1/2915 + 1/3135

= 1/2.(2/15 + 2/35 + 2/63 + 2/99 + ... + 2/2915 + 2/3135)

=1/2.( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/53.55 + 2/55.57)

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

=1/2.(1/3-1/57)

=1/2.6/19

=3/19

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

thanks các bn

\(A=\frac{4}{33}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{9.11}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(tắt 1 bước nha)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{9}-\frac{1}{11}\)

\(2A=\frac{1}{3}-\frac{1}{11}\)

\(2A=\frac{8}{33}\)

\(\Rightarrow A=\frac{4}{33}\)

Vậy A=_____________

49/303

\(A=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{49}{303}\)

ta có:

A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{999}\)

A=\(\frac{1}{3\text{x}5}+\frac{1}{5\text{x}7}+\frac{1}{7\text{x}9}+\frac{1}{9\text{x}11}+...+\frac{1}{?\text{x}?}\)

A=

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{9999}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+....+\frac{1}{99.101}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{101}\)

\(A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\) (vì tất cả các phân số khác ngoài 1/3 và 1/101 đều đã bị cộng với số đối với nó = 0)

A= 1/15+1/35+1/63+1/99+……+1/9999
=1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101  SAI