LH
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
YA
3
9 tháng 2 2018
A = 1 + 1/2.4 + 1/4.6 + ...... + 1/10.12
2A = 2 + 2/2.4 + 2/4.6 + ...... + 2/10.12
= 2 + 1/2 - 1/4 + 1/4 - 1/6 + ...... + 1/10 - 1/12
= 2 + 1/2 - 1/12 = 29/12
=> A = 29/12 : 2 = 29/24
P/S : Tham khảo nha
LT
Lương Thị Vân Anh
CTVHS
2 tháng 3 2023
\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}+\dfrac{1}{10\times12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{17}{12}\)
21 tháng 3 2023
=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)
21 tháng 3 2023
=1+12⋅4+14⋅6+...+110⋅121+12⋅4+14⋅6+...+110⋅12
=1+12(12−14+14−16+...+110−112)=1+12(12−14+14−16+...+110−112)
=1+12⋅512=1+524=2924
28 tháng 2 2018
A = 1 + 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10 + 1/10.12
2A = 2 + 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12
= 2 + 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10 + 1/10 - 1/12
= 2 + 1/2 - 1/12 = 29/12
=> A = 29/12 : 2 = 29/24
Tk mk nha
TD
1
5 tháng 4 2021
Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{29}{12}\)
hay \(A=\dfrac{29}{24}\)
5 tháng 5 2020
Bài 1:
gọi số đó là x
ta có : \(\frac{-1}{12}< x< \frac{-1}{2}\)
hay :
\(\frac{-1}{12}< x< \frac{-6}{12}\)
vậy \(x\in\left\{\frac{-2}{12};\frac{-3}{12};\frac{-4}{12};\frac{-5}{12}\right\}\)
Tính tổng tất cả các phân số có mẫu số là 12 là :
\(\frac{-2}{12}+\frac{-3}{12}+\frac{-4}{12}+\frac{-5}{12}=\frac{-14}{12}=\frac{-7}{6}\)
bài 2:
\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(2A=2+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)
\(2A=3-\frac{1}{12}\)
\(A=\left(\frac{35}{12}\right):2=\frac{35}{24}\)
6L
1
PN
8 tháng 7 2020
\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}\)
\(< =>A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(< =>2A=2+\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(< =>2A=2+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)
\(< =>2A=\frac{5}{2}-\frac{1}{12}=\frac{29}{12}\)
\(< =>A=\frac{29}{12}.\frac{1}{2}=\frac{29}{24}\)
LD
1
21 tháng 6 2016
\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}=\frac{5}{24}\)
Vậy tổng của các phân số đó bằng \(\frac{5}{24}\)
Mik đầu tiên đấy , ai click mik mik click lại
\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+\dfrac{1}{10.12}\)
\(2A=\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\)
\(2A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\)
\(2A=1-\dfrac{1}{12}\)
\(2A=\dfrac{11}{12}\)
\(A=\dfrac{11}{12}:2=\dfrac{11}{24}\)
\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{10\cdot12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{12}\right)=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)