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=

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

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

1/3-1/101

98/303

98/303 nhék

k mk nhak

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

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.9}\)

\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.9}\right)\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{1}{111.9}\)

= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{111}-\frac{1}{9}\)

\(=\frac{1}{3}-\frac{1}{9}\)

\(=\frac{2}{9}\)

Đặt phép tính cần tìm là A

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(2A=1-\dfrac{1}{13}\)

\(2A=\dfrac{12}{13}\)

\(A=\dfrac{6}{13}\)

\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)

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

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

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(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=_____________

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