A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{129.15}\)

A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{129.15}\)

A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{129}-\frac{1}{15}\)

A = \(\frac{1}{3}-\frac{1}{15}\)= \(\frac{5}{15}-\frac{1}{15}\)

A = \(\frac{4}{15}\)

CẢM ƠN LỜI GỢI Ý

kết quả đúng là 7/45 nhé

A=1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999

A= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ...+ 1/99x101

Ax2= 2/3x5 +  2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101

Ax2= 5-3/3x5 + 7-5/5x7 + 9-7/7x9 + 11-9/9x11 + ... + 101-99/99x101

Ax2=5/3x5 - 3/3x5 + 7/5x7 - 5/5x7 + 9/7x9 -7/7x9 + ... + 101/99x101 -99/99x101

Ax2=1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101

Ax2= 1/3 - 1/101

Ax2 = 98/303

A= 98/303 : 2

A=49/303

Ôi xin lỗi ! Đừng quên tick nha

A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15

A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15

A=1 - 1/15

A=1/14

bạn ơi: 

1/195 mới bằng 1/13.15 mà!!

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

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

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

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

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

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

A = 1/15 + 1/35 + 1/63 + 1/99 + ....... + 1/9999

A = 1/3 x 5 + 1/5 x 7 + 1/9 x7 + .........+ 1/99 x101

A = 1/2 x ( 1/3 -1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...... + 1/99 - 1/101

A = 1/2 x ( 1/3 - 1/99 )

A = 1/2 x 98/303

A = 49/303

A = 1/3.5 +1/5.7 + 1/7.9 + 1/9.11 + ... + 1/99. 101

= 1/2.(2/3.5+ 2/5.7 + 2/7.9 + ...+2/99.101)

= 1/2.(1/3 - 1/5 - 1/5 - 1/7 - 1/7 - 1/9  + .... +1/99 - 1/101

=1/2.(1/3 - 1/101)

=1/2 .98//303

=49/303

Dấu . là nhân đó nha bạn 

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

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

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

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

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

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

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

   \(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}\left(\frac{1}{99}-\frac{1}{101}\right)\)

   \(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

   \(=\frac{1}{2}.\frac{98}{303}\)

   \(=\frac{49}{303}\)

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