\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{98}{99}=\dfrac{1}{33}\cdot49=\dfrac{49}{33}\)

Lời giải:

$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$

$=1-\frac{1}{99}=\frac{98}{99}$

Đặt biểu thức = A

6A = 1x3x6 + 3x5x6+....+ 97x99x6

= 1x3x(1+5) + 3x5x(7-1) + 5x7x(9-3) +.....+ 97x99x(101-95)

= 1x3+1x3x5+3x5x7-1x3x5+5x7x9-3x5x7+.....+97x99x101-95x97x99

= 1x3+97x99x101

= 969906

=> A = 161651

k mk nha

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.....................  >.<