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!

=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

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

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

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

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

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

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

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

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

\(\Rightarrow A< 99\)

7/24 nha

k mình nhé

= 7/24

ai tk mình mình tk lại cho

S= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

S= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  1/5 -1/6 +1/6 - 1/7 + 1/7 - 1/8

S= 1/2 - 1/ 8 

S= 3/8

S= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8

  = 1/2 - 1/3 + 1/3 - ...+ 1/7 - 1/8

  = 1/2 - 1/8

  = 3/8