=\(\frac{1}{3\cdot5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11\cdot13}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{13}=\frac{5}{13}\) 

B=1/15+1/35+1/63+1/99+1/143

B=1/3.5+1/5.7+1/7.9+1/9.11+11.13    (khoảng cách từ 3-5;5-7;7-9;9-11;11-13 la 2)

Suy ra B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) (Ta gop -1/5+1/5;-1/7+1/7;-1/9+1/9;-1/11+1/11 bang 0)

B=1/2(1/3-1/43)=1/2.40/129=20/129

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

mình không biết nữa bằng bao nhiêu ấy nhỉ .......? .......? Sory ^.^

1/3 + 13/15 + 33/35 + 61/63 + 97/99

= 45/11 ( mình không tiện giải, để khi khác giải sau)

Chúc bạn may mắn!

= 45/11

mik làm biếng ghi lâu lắm bạn ạ !!!

k mik nhaaaaaaaaaaaaaaaaaaaaaa

98/303 tích nha mình giải cho

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

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

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

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

\(A=\frac{1}{2}\times\frac{98}{303}\)

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

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

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

2. A= \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) + ... + \(\frac{2}{99.101}\)

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

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

2.A= \(\frac{101}{303}\) - \(\frac{3}{303}\)

2.A= \(\frac{98}{303}\)

A  = \(\frac{98}{303}\) : 2

A  = \(\frac{49}{303}\)

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

43/303

Hình như sai đề rồi

ý đúng rồi

\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)

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

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

\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)

\(\frac{10}{11}.y=\frac{4}{3}\)

\(\Rightarrow y=\frac{22}{15}\)