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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\)

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

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

\(=\frac{8}{9}\)

115/126

2 / 3 + 2 / 15 + 2 / 35 + 2 / 63 + 2 / 99 
= 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7- 1/9 + 1/9 - 1/11 (công thức)
= 1 + 1/3 - 1/3 + 1/5 - 1/5 + 1/7 - 1/7 + 1/9 - 1/9 - 1/11 (giao hoán)
= 1 + (1/3 - 1/3) + (1/5 - 1/5) + (1/7 - 1/7) + (1/9 - 1/9) - 1/11
= 1 + 0 + 0 + 0 + 0 - 1/11 = 1 - 1/11 = 10/11

2/3+2/15+2/35+2/63+...+2/9999

=2/1.3+2/3.5+2/5.7+...+2/99x101

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

=1-1/101=100/101

\(=2\times\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

\(=2\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{99\times101}\right)\)

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

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

\(=2\times\frac{100}{101}\)

\(=\frac{200}{101}\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{99.101}\)

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

\(=1-\frac{1}{101}=\frac{100}{101}\)

Giải:

\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\) 

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

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

\(=\dfrac{1}{1}-\dfrac{1}{13}\) 

\(=\dfrac{12}{13}\) 

Chúc em học tốt!

2/3+2/15+2/35+2/63+2/99+2/143

=2(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)

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

=2(1-1/13)

=2.12/13=24/13

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{1}{5.7}+....+\frac{2}{99.101}\)

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

\(=1-\frac{1}{101}+0+0+...+0\)

\(=\frac{100}{101}\)

Bài làm :

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

\(=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)

\(=2\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}\right)\)

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

\(=2\times\left(1-\frac{1}{11}\right)\)

\(=2\times\frac{10}{11}\)

\(=\frac{20}{11}\)

Học tốt nhé

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

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\)

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

\(=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)

         Bài làm :

Ta có :

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

\(=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)

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

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

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

2/15+2/35+2/63+.....+2/9603=2/3.5+2/5.7+2/7.9+...2/97.99

=1/3-1/5+1/5-1/7+...+1/97-1/99

=1/3-1/99

=33/99-1/99

=32/99

\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{9603}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

2/15 + 2/35 + 2/63 + 2/99 + 2/143 + 2/195

\(=2\times\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\right)\)

= \(2\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=2\times\left(\dfrac{1}{3}-\dfrac{1}{15}\right)\)

\(=2\times\dfrac{4}{15}\)

\(=\dfrac{8}{15}\)

            A=12/15 + 28/315

            A=8/9 

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(=\frac{3-1}{3}+\frac{15-1}{15}+\frac{35-1}{35}+\frac{63-1}{63}+\frac{99-1}{99}\)

\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)

\(=5+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(=5+\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(=5+\frac{5}{11}=\frac{60}{11}\)