Ta có:

    A=5/15+5/35+5/63+5/99+...+5/2915

=>A=5/3.5+5/5.7+5/7.9+5/9.11+...+5/53.55

=>A=5/2.(2/3.5+2/5.7+2/7.9+2/9.11+...+2/53.55)

=>A=5/2.(2/3-2/5+2/5-2/7+2/7-2/9+2/9-2/11+...+2/53-2/55)

=>A=5/2.(2/3-2/55)

=>A=5/2.104/165

=>A=52/33

Vậy A=52/33

OK!

thank

\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{19\times21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{7}{21}-\frac{1}{21}\)

\(=\frac{6}{21}\)

Rút gọn kết quả là \(\frac{2}{7}\), k mk nha mk trả lời đầu tiên đó

\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+....+\frac{2}{399}\)

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{6}{21}=\frac{2}{7}\)

\(=\frac{4.4}{3.5}.\frac{5.5}{4.6}......\frac{20.20}{19.21}\)

\(=\left(\frac{4.5...20}{3.4....19}\right).\left(\frac{4.5...20}{5.6....21}\right)\)

\(=\frac{20}{3}.\frac{4}{21}\)

\(=\frac{80}{63}\)

\(=\frac{4.4}{3.5}.\frac{5.5}{4.6}.....\frac{20.20}{19.21}\)

=\(\left(\frac{4.5...20}{3.4...19}\right).\left(\frac{4.5.....20}{5.6....21}\right)\)

=\(\frac{20}{3}.\frac{4}{21}\)=\(\frac{80}{63}\)

hok tốt

\(\frac{2}{7}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

chi tiết đc hơn ko bạn

giúp mình ik nhá

\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)

b)

\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)

\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)

=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)

\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)

\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)

=> \(B=\frac{140}{69}:2=\frac{70}{69}\)

a) SỬA LẠI ĐỀ : A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{6480}\)

       = \(5.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

       =\(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

        = \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

        = \(5.\left(1-\frac{1}{81}\right)\)

        = \(5.\frac{80}{81}\)

        = \(\frac{400}{81}\)

b) B = \(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+...+\frac{7}{483}\)

        = \(\frac{7}{3.5}+\frac{7}{5.7}+\frac{7}{7.9}+...+\frac{7}{21.23}\)

        = \(\frac{7}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\frac{20}{69}\)

        = \(\frac{70}{69}\)

Ta có \(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

\(=2.\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{2}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{2}{7}\)

Vậy \(B=\frac{2}{7}\)

ta có 

A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{19.21}\)

\(=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{21}\right)\)

=\(\frac{4}{7}\)

\(A=\)\(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{19.21}\)\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

\(y=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

mình thắc mắc quy luật của phép tính trên là gì : 15 -> 35 -> 63 ... -> 399 ?

Quy luật la cac so o duoi mau cach nhau 2 don vi ( cach nhau mot so bang tu):

2/15=2/3.5 ( 3 va 5 cach nhau 2don vi)

2/35=2/5.7 ( 5 va 7 cach nhau 2 don vi)

.........

(Cai nay minh gui cho ban Duy Thanh nhe!!!!!!!!!!!!!!!!!!!)

Ta có:

\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

   \(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

   \(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

các bạn.