= 2/3×5 + 2/ 5×7 + 2/ 7×9 + 2/ 9×11 + ...... + 2/ 23×25+ 2 / 25×27

=1/3 - 1/5 + 1/5 -1/7 + 1/7 -1/9 +1/9 -1/11+.......+1/23- 1/25 + 1/25 -1/27

= 1/3 - ( 1/5+1/5 -1/7 +1/7 - 1/9 +1/9-1/11 +........+ 1/23 -1/25 + 1/25 )-1/27

=1/3 - 1/27= 9/27 - 1/27= 8/27

11/15 nha

BẰNG 11/15 NHA BẠN

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

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

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn

=2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
= 1/3 - 1/13
= 13/39 -3/39
=10/39 

bằng 10/39 nha

chúc bạn học tốt

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\\ =\dfrac{1}{3}-\dfrac{1}{13}\\ =\dfrac{13}{39}-\dfrac{3}{39}\\ =\dfrac{10}{39}\)

NHỚ TÍNH ĐÚNG VỚI CHO COIN NHA

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

= 2/1×3 + 2/3×5 + 2/5×7 + 2/7×9 + 2/9×11

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

= 1 - 1/11

= 10/11

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

=4/5+2/15+2/63+2/99

=14/15+2/63+2/99

=304/315+2/99

=1138/1155

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

\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

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

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

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

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

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

= \(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}\)

= \(\frac{3}{1\times3}-\frac{1}{1\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+...+\frac{11}{9\times11}-\frac{9}{9\times11}\)

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

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

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

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\)

\(=\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+\frac{9}{7.9}-\frac{7}{7.9}+\frac{11}{9.11}-\frac{9}{9.11}\)

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

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

\(bang 12.603\)

Làm như nào zậy