\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

A=1/1.2+1/2.3+...+1/49.50

  =1-1/2+1/2-1/3+...+1/49-1/50

  =1-1/50

  =49/50

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(\Rightarrow A=\frac{49}{50}\)

Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
 1) Ta có:    \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)

                                                                             \(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)

                                                                             \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}

Ta có : 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vậy \(A=\frac{49}{50}\)

Chúc bạn học tốt ~ 

A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

A= \(1-\frac{1}{50}\)

A= \(\frac{49}{50}\)

A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

A= \(1-\frac{1}{50}\)

A= \(\frac{50}{50}-\frac{1}{50}\)

A= \(\frac{49}{50}\)

Vậy A= \(\frac{49}{50}\)

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{97}{48^2.49^2}+\frac{99}{49^2.50^2}\)

\(\Leftrightarrow\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{97}{2304.2401}+\frac{99}{2401.2500}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{2304}-\frac{1}{2401}+\frac{1}{2401}-\frac{1}{2500}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2500}=\frac{2499}{2500}< 1\left(đpcm\right)\)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)

ta thấy 1/(1*2)-1/(2*3)=1/3=2*1/(1*2*3)

do đó A=1/2*{[1/(1*2)-1/(2*3)+[1/(2*3)-1/(3*4)]+.....+[1/(48*49)-1/(49*50)]} 

            =1/2*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+.....+1/(48*49)-1/(49*50)]

            =1/2*[1/(1*2)-1/(49*50)]

            =1/2*(1/2-1/2450)

             =1/2*612/1225

            =306/1225

A= 306/1225

pạn cho mk cách giải dc k?

Gọi tổng trên là A

A = 1/1.2.3 + 1/2.3.4 +......+ 1/48.49.50

2A = 2/1.2.3 + 2/2.3.4 +.......+ 2/48.49.50

2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +.......+ 1/48.49 - 1/49.50

2A = 1/1.2 - 1/49.50

2A = 1224/2450

A = 612/2450 = 306/1225

\(A=2.3+3.4+4.5+...+49.50\)

\(3A=2.3.3+3.4.3+4.5.3+...+49.50.3\)

\(3A=2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+49.50.\left(51-48\right)\)

\(3A=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50\)

\(3A=-1.2.3+49.50.51\)

\(3A=-6+48450\)

\(3A=48444\)

\(A=\frac{48444}{3}\)

\(A=16148\)

Chúc bạn học tốt ~ 

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{25.26.27}\)

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{25.26.27}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{25.26}-\frac{1}{26.27}\)

\(2B=\frac{1}{1.2}-\frac{1}{26.27}\)

\(2B=\frac{1}{2}-\frac{1}{702}\)

\(2B=\frac{175}{351}\)

\(B=\frac{175}{251}:2\)

\(B=\frac{175}{502}\)

Chúc bạn học tốt ~