A= 1/2.5+1/5.8 + 1/8.11 +........ 1/152.155
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25 tháng 7 2015
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{152.155}\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{152}-\frac{1}{155}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{155}\right)\)
\(A=\frac{1}{3}.\frac{153}{310}\)
\(A=\frac{51}{310}\)
16 tháng 3 2016
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
NQ
3
TD
21 tháng 3 2016
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
KV
11 tháng 11 2020
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{152.155}\)
\(=\frac{1}{3}.\frac{5-2}{2.5}+\frac{1}{3}.\frac{8-5}{5.8}+\frac{1}{3}.\frac{11-8}{8.11}+...+\frac{1}{3}.\frac{155-152}{152.155}\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{152}-\frac{1}{155}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{155}\right)\)
\(=\frac{1}{3}.\frac{153}{310}\)
\(=\frac{51}{310}\)
Vậy A \(=\frac{51}{310}\)
VA
18 tháng 1 2022
tham khảo ở đây nha
https://olm.vn/hoi-dap/detail/222956295982.html
18 tháng 1 2022
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{150}-\dfrac{1}{153}\right)\)
\(=\dfrac{1}{3}.\dfrac{151}{306}=\dfrac{151}{918}\)
2 tháng 5 2018
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 = \(\frac{8}{49}\)
WH
2 tháng 5 2018
\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)
\(\Rightarrow3A=3\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\right)\)
\(\Rightarrow3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}\)
\(\Rightarrow3A=\frac{24}{49}\)
\(\Rightarrow A=\frac{24}{49}:3\)
\(\Rightarrow A=\frac{8}{49}\)
Vậy \(A=\frac{8}{49}\)
9 tháng 6 2019
Tham khảo tại link này nhé : https://olm.vn/hoi-dap/detail/79256477545.html
hoặc vô câu hỏi tương tự cx có đó
9 tháng 6 2019
\(A=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{95.98.}\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{24}{49}\)
\(A=\frac{8}{49}\)
MA
1
KQ
21 tháng 2 2023
`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`
3 tháng 5 2018
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)
3 tháng 5 2018
\(=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=3.\frac{24}{49}\)
\(=\frac{72}{49}\)
