\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)

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

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

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

=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

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

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

A=  2 + 6 + 12 + 20 + ...... + 9702 + 9900

A = 1.2 + 2.3 + 3.4 + ......... + 98 . 99 + 99.100

3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

3A = 1.2.3 + 2.3.(4-1) + ....+ 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ..... + 99.100.101 - 98.99.100

3A = 99 . 100 . 101

A = 99 . 100 . 101 : 3

A = 333300

A=1.2+2.3+3.4+4.5+....+.....

3A=.....

Bạn biets làm rồi đúng ko

Tích mk nha hùng

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

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

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

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

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

\(=\dfrac{33}{99}-\dfrac{1}{99}=\dfrac{32}{99}\)

CÓ AI GIÚP MK VS NHA