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

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

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

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

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

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

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

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

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

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

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

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

   \(=\frac{1}{2}.\frac{98}{303}\)

   \(=\frac{49}{303}\)

Dấu chấm(.) ở cấp hai là dấu nhân (x)

làm rồi

2/5

tick nha

= 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

tick cai nao cho dung

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

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

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

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

\(=\frac{1}{2}.\frac{98}{303}=\frac{49}{303}\)

49/303 nha bạn

Kb với mình rồi mình giải kĩ cho

@@@@@###

1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11)+...  + 1/(99x101)

(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/99-1/101) : 2

(1/3 - 1/101) : 2 =  98/303  :  2

49/303

Bạn đưa về dãy tổng

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

Có thể tính nhanh vì đây là dãy đặc biệt 

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

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

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

Sau khi lược bỏ các phân số ( phân số cộng với nhau bằng 0 coi như là không cộng)

Ta còn : \(\frac{1}{3}-\frac{1}{101}\)=\(\frac{98}{303}\)

Đáp số: \(\frac{98}{303}\)

\(\frac{98}{303}\)

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

A     = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)

Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)

Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101

Ax2 = 1/3 – 1/101 = 98/303

A    =  98/303 : 2

A    =  49/303