A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/255

A = 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 +...+ 1/15x17

2A = 2/3x5 + 2/5x7 + 2/7x9 + .... + 1/15x17

2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + .... + 1/15 - 1/17

2A = 1/3 - 1/17 = 14/51

A = 14/51 : 2 = 7/51

ta tách biểu thức trên thành 1/ 3*5+1/ 5*7+1/ 7*9+1/ 9*11+...+1/ 15*17

từ biểu thúc trên có:

1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/15-1/17

rút gọn còn  1/3-1/17

còn lại bạn tự tính nha

zin cho cách giải lun nha bạn crityamo the anh

A =............=  867

đề sai sao làm được????????????????

A=1/15+1/35+1/63+1/99+1/143+1/195+1/255+1/323

=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19

=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19

=> 2A = 1/3 - 1/19

=> 2A = 16/57 => A = 16/57 : 2 = 8/57

=>=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19

=>=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19

=> 2A = 1/3 - 1/19

=> 2A = 16/57 => A = 16/57 : 2 = 8/57

1/15+1/35+1/63+1/99+1/143+1/195+1/255+1/323

=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15+1/15-1/17+1/17-1/19

=2/3-2/5+2/5-2/7+2/7-2/9+2/9-2/11+2/11-2/13+2/13-2/15+2/15-2/17+2/17-2/19

=1/3-1/19

2/9

=1/3*5+1/5*7+1/7*9+...+1/99*101

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

=1/3-1/101

=98/303

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

= 1/(3x5) + 1/(5x7) + 1/(7x9) + ... + 1/(99x101)

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

= (1/3 - 1/101) : 2 

= 98/303 : 2

= 49/303

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

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

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

\(a=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{49}{303}\)

1/3 + 1/15 + 1/35 + ...... + 1/9999

= 1/1.3+ 1/3.5+1/99.101

= 1/2.( 2/1.3+ 2/3.5+..+2/99.101)

= 1/2.(1-1/3+1/3-1/4+.....+ 1/99-1/101)

= 1/2.( 1-1/101)= 1/2.100/101

= 50/101

tk đi rồi mk tk lại......

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

A=1/2 [1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13]

A=1/2.10/39

A=5/39

ai tk mình

mình tk lại

=5/39

 tk mk đi mk tk cho hứa đó

Đặt \(B=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)

\(\Leftrightarrow B=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)

\(\Leftrightarrow2B=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{13}\right)=1-\frac{3}{13}=\frac{10}{13}\)

\(\Leftrightarrow A=1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}=1+\frac{10}{13}=\frac{23}{13}\)

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

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

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

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

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

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

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

B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13

   = 1/2 * ( 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13)

   = 1/2 * ( 1/3 - 1/5 + 1/5 -1/7 + ...+ 1/11 - 1/13)

   = 1/2 * ( 1/3 - 1/11)

   = 1/2 * 8/33

   = 4/33

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

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

\(B=\frac{1}{2}.\frac{12}{39}\)

\(B=\frac{2}{13}\)

\(\Rightarrow2B=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(\Rightarrow\)\(2B=\frac{1}{3}-\frac{1}{13}\)\(=\frac{13}{39}-\frac{3}{39}=\frac{10}{39}\)\(\Rightarrow B=\frac{10}{\frac{39}{2}}=\frac{10}{39\cdot2}=\frac{5}{39}\)