T= 1 - 1/2 + 1/2 - 1/3 + ......+ 1/99 - 1/100

  = 1 - 1/100

  = 99/100

\(t=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(t=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(t=1-\frac{1}{100}=\frac{99}{100}\)

Vậy \(t=\frac{99}{100}\)

dấu . hiểu là phép nhân nhé

tớ cũng nghĩ là 99/100

ta có : t = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/98.99 + 1/99.100

=> t = 1/1 - 1/2 + 1/2 - 1/3 + .... + 1/99 - 1/100

=> t = 1 - 1/100

=> t = 99/100

T=1/1x2+1/2x3+1/3x4+....................+1/98x99+1/99x100

T=1-1/2+1/2-1/3+..............+1/98-1/99+1/99-1/100

T=1-1/100

T=99/100

T=1/2+1/6+1/12+..............+1/9702+1/9900

T=1/1x2+1/2x3+1/3x4+...........+1/98x99+1/99x100

T=1-1/2+1/2-1/3+1/3-1/4+.........+1/98-1/99+1/99-1/100

T=1-1/100

T=99/100

           Vậy T=99/100

Giải :

Đặt : A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9702}+\frac{1}{9900}\)

\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}=\frac{99}{100}\)

99/100 duyệt đi

T = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\)

T = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\)

T = \(\frac{1}{1}-\frac{1}{100}=\)

T = \(\frac{99}{100}\)

T là 99/100 . Đúng 100% luôn nhé .

\(\frac{1}{2}+\frac{1}{6}+............+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

99

100

Có: \(A=\frac{1}{2}+\frac{5}{6}+...+\frac{9899}{9900}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{9900}\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=99-\left(1-\frac{1}{100}\right)\)

\(=99-\frac{99}{100}< 99\)

\(\Rightarrow A< 99\)