$\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}=\frac{2}{3.5}+\frac{2}{5.7}+$\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}$

=$\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{13}$

=$\frac{1}{3}-\frac{1}{13}=\frac{10}{39}$

Vậy....

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

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

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

\(S=\dfrac{1}{3}-\dfrac{1}{13}\)

\(S=\dfrac{13}{39}-\dfrac{3}{39}\)

\(S=\dfrac{10}{39}\)

Vậy \(S=\dfrac{10}{39}\)

11/15 nha

BẰNG 11/15 NHA BẠN

mi tích tau tau tích mi xong tau trả lời nka việt nam nói là làm

(tính bằng máy tính)

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

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

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

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

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

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

\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

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

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

\(A=\frac{1}{3}-\frac{1}{13}\)

\(A=\frac{10}{39}\)

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

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

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

\(=\frac{1}{3}-\frac{1}{13}\)

\(=\frac{10}{39}\)

_Hok tốt_

!!!

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

\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

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

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

\(A=\frac{2}{3}+\frac{2}{15}+...+\frac{2}{143}\)

\(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{11\cdot13}\)

\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)

\(A=1-\frac{1}{13}=\frac{12}{13}\)

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

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

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

\(=1-\frac{1}{13}\)

\(=\frac{12}{13}\)

\(\frac{2}{6}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

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

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(=1-\frac{1}{13}\)

\(=\frac{12}{13}\)