A = 2/3 x 5+ 2/5 x 7 + 2/7 x 9 + ... + 2/97 x 99

A = 2/3 - 2/99

A = 64/99

A=2/3x5 + 2/5x7 + 2/7x9 + ......+ 2/97x99 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 
A= 1/3 - 1/99 = 96/3.99 = 32/99 

        Đ/S:tích

A = 2/3 x 5 + 2/ 5 x 7 + 2/7 x 9 + ... + 2/97 x 99

A = 1/3 - 1/99

A = 32/99

2/3x5 + 2/5x7 + 2/7x9 + ......+ 2/97x99 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 
= 1/3 - 1/99 = 96/3.99 = 32/99 

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)

=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

=\(\frac{32}{99}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{33}{99}-\frac{1}{99}\)

\(=\frac{32}{99}\)

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

 Đúng 1 

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

\(\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)

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

\(\Rightarrow C=\frac{1}{3}-\frac{1}{99}\)

\(\Rightarrow C=\frac{32}{99}\)

C = 2/3x5 + 2/5x7 + 2/7x9 + ... + 2/97x99

C = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99

C = 1/3 - 1/99

C = 32/99

C=1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99

C=1/3-1/99

C=296/99

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

sai đề

\(\frac{1}{1x2} +(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9} +\frac{2}{9x11})\)

\(=\frac{1}{1x2} + (\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)

\(=\frac{1}{1x2}+(\frac{1}{3}-\frac{1}{11})\)

\(=\frac{1}{1x2} +\frac{10}{33}\)

\(=\frac{1}{2} + \frac{10}{33} = \frac{33}{66}+\frac{20}{66}\)

\(=\frac{53}{66}\)

đúng mà

B=1x3+3x5+5x7+7x9+...+95x97+97x99

= 1.(1+2)+3.(3+2)+5.(5+2)+....+95.(95+2)+97.(97+2)

= 1²+1.2+3²+3.2 +5²+5.2+...+95²+95.2+ 97²+97.2

= (1²+3² +5²+...+95²+ 97²)+(1.2+3.2 +5.2+...+95.2+97.2)

= (1²+3² +5²+...+95²+ 97²)+ 2.(1+3 +5+...+95+97)

Đặt : A = 1²+3² +5²+...+95²+ 97² 

C =1+3 +5+...+95+97  

    tính A và C (tìm câu hỏi tương tự hình như anh thấy họ làm rồi đấy) sau đó thay vào tính B