\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)

=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)

=> \(A=\frac{451}{8120}\)

Ta có \(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\dfrac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

Áp dụng:

\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)

\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)

\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)

Lại phải giải hết 
Gọi dãy số trên là A
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{200.201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-.....+\frac{1}{200.201.202}-\frac{1}{201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{201.202.203}\)(chỗ này lm hơi tắt tí )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)

Giups mk với...

chờ tối nha chớ giờ giải là khỏi đi học lun

bạn vào đây nè: http://olm.vn/hoi-dap/question/601925.html

\(A=\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+\dfrac{1}{3\cdot4\cdot5\cdot6}+....+\dfrac{1}{9\cdot10\cdot11\cdot12}\)

\(3A=\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+\dfrac{3}{3\cdot4\cdot5\cdot6}+...+\dfrac{3}{9\cdot10\cdot11\cdot12}\)

\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{9\cdot10\cdot11}-\dfrac{1}{10\cdot11\cdot12}\)\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{10\cdot11\cdot12}\)

\(A=\dfrac{1}{2}-\dfrac{1}{440}\)

\(A=\dfrac{219}{440}\)

Nhận xét: 1/1.2.3 - 1/2.3.4 = 3/1.2.3.4, 1/2.3.4 - 1/3.4.5 =3/2.3.4.5,...,1/27.28.29 - 1/28.29.30

Gọi biểu thức phải tính bằng A,ta tính được:

3A=1/2.3 - 1/28.29.30 = 4059/28.29.30

vậy A = 1353/8120

Ket quả cua mình là 451/8120

1353/8120

P = 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6 + ... + 1/97.98.99.100

P = 1/1-1/2-1/3-1/4+1/2-1/3-1/4-1/5 +....+1/97-1/98-1/99-1/100

P = 1/1-1/100

P = 99/100

Tính giá trị biểu thức P.3.98.99

Cái đó thì bạn tự tính cũng dc dễ mak

a,\(\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{11.13}\)

=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{11}-\frac{1}{13}\)

=\(\frac{1}{3}-\frac{1}{13}\)

=\(\frac{10}{39}\)

b,Đặt A=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.............+\frac{1}{27.28.29.30}\)

3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...........+\frac{3}{27.28.29.30}\)

3A=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.............+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

3A=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

3A=\(\frac{1}{6}-\frac{1}{24360}\)

3A=\(\frac{1353}{8120}\)

A=\(\frac{451}{8120}\)