=1-1/50=\(\frac{49}{50}\)

Tách: 1/1.2=1-1/2;  1/2.3=1/2-1/3; ....; 1/49.50=1/49-1/50

Và rút gọn các số liền kề thì còn lại kết quả

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

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

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

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

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

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

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

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

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

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

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

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

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

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= \(\frac{49}{50}\)

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)

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

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

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

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

\(A< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{49.50.51}.\)

\(2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{49.50.51}\)

\(2A< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{51-49}{49.50.51}\)

\(2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)

\(2A< \frac{1}{2}-\frac{1}{50.51}< \frac{1}{2}\Rightarrow A< \frac{1}{4}< \frac{1}{2}\)

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

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

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

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

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?