A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\) + ... + \(\dfrac{1}{97\times99}\)

A = \(\dfrac{1}{2}\) x (\(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\) + \(\dfrac{2}{7\times9}\) + ... + \(\dfrac{2}{97\times99}\))

A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}-\dfrac{1}{9}\) + ... + \(\dfrac{1}{97}\) - \(\dfrac{1}{99}\))

A = \(\dfrac{1}{2}\) x ( \(\dfrac{1}{3}\) - \(\dfrac{1}{99}\))

A  = \(\dfrac{1}{2}\times\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)

A = \(\dfrac{1}{2}\times\dfrac{32}{99}\)

A = \(\dfrac{16}{99}\)

B=3×53​+5×73​+7×93​+...+48×503​ 

\(B = \frac{3}{2} \times \left(\right. \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \frac{2}{7 \times 9} + . . . + \frac{2}{48 \times 50} \left.\right)\) 

\(B = \frac{3}{2} \times \left(\right. \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{48} - \frac{1}{50} \left.\right)\) 

\(B = \frac{3}{2} \times \left(\right. \frac{1}{3} - \frac{1}{50} \left.\right)\)

\(B = \frac{3}{2} \times \frac{47}{150}\) 

\(B = \frac{47}{100}\) 

quá dễ :

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

A=3/2.(1/3-1/5+1/5-1/3+...+1/97-1/99)

A=3/2.(1/3-1/99)

A=3/2.32/99

A= 16/33 

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

A = 3/2 . (1/3 - 1/5 + 1/5 - 1/3 + ... + 1/97 - 1/99)

A = 3/2 . (1/3 - 1/99)

A = 3/2 . 32/99

A = 16/33 

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 

Ta có \(6B=1\times3\times6+3\times5\times6+...+97\times99\times6\)

\(=1\times3\times\left(5+1\right)+3\times5\times\left(7-1\right)+5\times7\times\left(9-3\right)+...+97\times99\times\left(101-95\right)\)

\(=1\times3\times5+1.3+3\times5\times7-3\times5\times1+...-97\times99\times95\)

\(=97\times99\times101+3\)

\(\Rightarrow B=\frac{97\times99\times101+3}{6}=161651\)

6B=1x3x6+3x5x6+5x7x6+.....+97x99x6

6B=1x3x(5+1)+3x5x(7-1)+....+97x99x(102-95)

6B=1x3x5+1x3+3x5x7-3x5+....+97x99x101-95x97x99

6B=1x3x97x99x101

6B=969906

=>B=161651

6B = 1x3x6 + 3x5x6 + 5x7x6 +...+ 99x101x6

     = 1x3(5+1)+3x5(7-1)+...+97x99(101-95)

     = 1x3x5+1x3+3x5x(7-1)+...-97x99x95

     =97x99x101+3       rồi tự làm.....................  >.<

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 

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

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

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

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

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

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

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

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

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

=1/5-1/7 + 1/7 - 1/9 + 1/9 - 1/11+....+1/97-1/99

=1/5 -1/99

=....

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