\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{152.155}\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{152}-\frac{1}{155}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{155}\right)\)

\(A=\frac{1}{3}.\frac{153}{310}\)

\(A=\frac{51}{310}\)

nhầm : 51/310

   1/2.5+1/5.8+1/8.11+...+1/152.155

=1/3(3/2.5+3/5.8+3/8.11+...+3/152.155

=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/152-1/155)

=1/3(1/2-1/155)

=1/3(155/310-2/310)

=1/3.153/310=51/310

Kết quả:51/310

A= 1/2.5+1/5.8+1/8.11+...+1/152.155

A= 1/2-1/5+1/5-1/8+...+1/152-1/155 loại bỏ các số giống nhau ta được:

A=1/2-1/155=153/310

Vậy A=153/310

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{152.155}\)

\(=\frac{1}{3}.\frac{5-2}{2.5}+\frac{1}{3}.\frac{8-5}{5.8}+\frac{1}{3}.\frac{11-8}{8.11}+...+\frac{1}{3}.\frac{155-152}{152.155}\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{152}-\frac{1}{155}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{155}\right)\)

\(=\frac{1}{3}.\frac{153}{310}\)

\(=\frac{51}{310}\)

Vậy A \(=\frac{51}{310}\)

A= 1/2.5+1/5.8+1/8.11+...+1/152.155

A= 1/2-1/5+1/5-1/8+...+1/152-1/155 loại bỏ các số giống nhau ta được:

A=1/2-1/155=153/310

Vậy A=153/310

tham khảo ở đây nha

https://olm.vn/hoi-dap/detail/222956295982.html

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{150}-\dfrac{1}{153}\right)\)

\(=\dfrac{1}{3}.\dfrac{151}{306}=\dfrac{151}{918}\)

3A = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\)

3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)

3A=\(\frac{1}{2}-\frac{1}{98}\)

3A=\(\frac{98}{196}-\frac{2}{196}\)=\(\frac{96}{196}=\frac{24}{49}\)

A=\(\frac{24}{49}:3=\frac{24}{49}.\frac{1}{3}=\frac{8}{49}\)

Vậy A = \(\frac{8}{49}\)

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)

\(\Rightarrow3A=3\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\right)\)

\(\Rightarrow3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}\)

\(\Rightarrow3A=\frac{24}{49}\)

\(\Rightarrow A=\frac{24}{49}:3\)

\(\Rightarrow A=\frac{8}{49}\)

Vậy \(A=\frac{8}{49}\)

\(A=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=3.\frac{24}{49}\)

\(=\frac{72}{49}\)

Tham khảo tại link này nhé : https://olm.vn/hoi-dap/detail/79256477545.html

hoặc vô câu hỏi tương tự cx có đó

\(A=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{95.98.}\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\frac{24}{49}\)

\(A=\frac{8}{49}\)

Trong câu hỏi tương tự có bạn ạ tham khảo nhé

~Hok tốt~

=21​−51​+51​−71​+....+951​−981​

=12−198=21​−981​tự làm tiếp

A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98

A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )

A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )

A = 1/3 . ( 1/2 - 1/98 )

A = 1/3 . 24/49

A = 8/49 tick cho tui

`D=1/2.5+1/5.8+1/8.11+...+1/1979.1982`

`=3/3(1/2.5+1/5.8+1/8.11+...+1/1979.1982)`

`=1/3(3/2.5+3/5.8+3/8.11+...+3/1979.1982)`

`=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/1979-1/1982)`

`=1/3(1/2-1/1982)`

`=1/3(991/1982-1/1982)`

`=1/3 . 495/991`

`=165/991`

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)

\(=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=3.\frac{24}{49}\)

\(=\frac{72}{49}\)

mk lm sai các bn đừng tk sai nha! xin m.n đó, mk chỉ chưa đọc kĩ đề thôi nha!