\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\\ =\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{98\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

1/3.4 + 1/4.5 + ...+1/98.99

= 1/3-1/4+1/4-1/5+...+1/98-1/99

= 1/3-1/99= 32/99

\(\dfrac{32}{99}\)

Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

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

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

Giải:

1/12+1/20+1/30+...+1/9702

=1/3.4+1/4.5+1/5.6+...+1/98.99

=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99

=1/3-1/99

=32/99

Chúc bạn học tốt!

Ta sử dụng t/c sau:

`1/(a(a+1))=1/a-1/(a+1)`

`=>1/12+1/20+1/30+...+1/9072`

`=1/(3.4)+1/(4.5)+....+1/(98.99)`

`=1/3-1/4+1/4-1/5+....+1/98-1/99`

`=1/3-1/99`

`=32/99`

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

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

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

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

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

99

100

=1 phần 3*4+1 phần 4*5+1 phần 5*6+...+1 phần 98*99

=1 phần 3-1 phần 4+ 1 phần 4- 1 phần 5+...+1 phần 98-1 phần 99

=1 phần 3- 1 phần 99                      =32 phần 99

\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{33-1}{99}=\dfrac{32}{99}\)

\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)

= \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\)

= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

= \(\dfrac{1}{3}-\dfrac{1}{99}\)

= \(\dfrac{33}{99}-\dfrac{1}{99}\)

= \(\dfrac{32}{99}\)

Kết quả là 99/100

 bằng 99/100 nha bạn

thi rồi, đúng 99,9% đó

bằng 99 /100

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

T= 1/1.2+1/2.3+1/3.4+...+1/99.100

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

T=1- 1/100

T= 99/100

đúng cho mình nha bạn

Bài này đơn giản mà bạn

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

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

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

T=99/100