4/2.5 + 4/5.8 + 4/8.11 + ... + 4/ 302 . 305
4/2.5+4/5.8+4/8.11+...+4/62.65
\(=\frac{3}{4}\cdot\left(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{62.65}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\)
\(=\frac{1}{2}-\frac{1}{65}\)
\(=\frac{63}{130}\)
Đặt A=4/2.5+4/5.8+4/8.11+...+4/62.65.Ta có A=4.(1/2.5+1/5.8+1/8.11+...1/62.65)=4/3.(3/2.5+3/5.8+3/8.11+...+3/62.65) =4/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+3/62-3/65)=4/3.(1/2-1/65)=4/3.63/130=42/56 Vậy A=42/56
B= 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14
B=?
bằng 3.141939849 kết bạn với mình nhé
Ta có : B = 4 / 2. 5 + 4 / 5 . 8 + 4 / 8 . 11 + 4 / 11 . 14
= 4 ( 1 / 2 . 5 + 1 / 5 . 8 + 1 / 8 . 11 + 1 / 11 . 14 )
= 4 . [ 1/3 ( 1 /2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/ 11 + 1/11 - 1/14 )]
= 4 / 3 ( 1/ 2 - 1/14 )
= 4/3 . 3/7
= 4 / 7
Vậy B = 4 / 7
B = \(\frac{4}{2\times5}+\frac{4}{5\times8}+\frac{4}{8\times11}+\frac{4}{11\times14}\)
B = \(\frac{4}{10}+\frac{4}{40}+\frac{4}{88}+\frac{4}{154}\)
B = \(\frac{1}{73}\)\(\left(\frac{4}{292}\right)\)
M= 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20
M = 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20
M= 4/3 . (1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
M= 4/3 . (1/2 - 1/20)
M= 4/3 . (10/20 - 1/20)
M= 4/3 . 9/20
M= 3/5
k nha
Bài 7: A=\(\dfrac{4}{2.5}\)+\(\dfrac{4}{5.8}\)+\(\dfrac{4}{8.11}\)+...+\(\dfrac{4}{65.68}\)
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
Tính A= 4/2.4+4/5.8+8.11+...+4/65.68
( 4/2.5=4 trên 2.5 )
Ai lm đc nhớ nghi lời, cách giải đầy đủ~
Cảm ơn!!!
Sửa đề:
\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)
\(A=4.\dfrac{33}{68}\)
\(A=\dfrac{33}{17}\)
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
Tính
A=4/2.5+4/5.8+4/8.11+...+4/65.68
\(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+\dfrac{4}{8\cdot11}+...+\dfrac{4}{65\cdot68}\\ =\dfrac{4}{3}\cdot\dfrac{3}{2\cdot5}+\dfrac{4}{3}\cdot\dfrac{3}{5\cdot8}+\dfrac{4}{3}\cdot\dfrac{3}{8\cdot11}+...+\dfrac{4}{3}\cdot\dfrac{3}{65\cdot68}\\ =\dfrac{4}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{65\cdot68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\dfrac{33}{68}\\ =\dfrac{11}{17}\)
Tính tổng :
M= 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20
3/4 . M = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20
3/4 . M = 1/2 - 1/20
3/4 .M = 9/10
M = 9/10 : 3/4
M = 3/5
Vậy M = 3/5
Nhớ tk m nha
Tính : \(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\) + \(\dfrac{4}{8.11}\) + ... + \(\dfrac{4}{65.68}\)
7A = \(\dfrac{4.3}{2.5}\) + \(\dfrac{4.3}{5.8}\) + \(\dfrac{4.3}{8.11}\) + ... + \(\dfrac{4.3}{65.68}\)
7A = 4 (\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{65.68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + ... + \(\dfrac{1}{65}\) - \(\dfrac{1}{68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
7A = 4 . \(\dfrac{33}{68}\) = \(\dfrac{33}{17}\)
A = \(\dfrac{33}{17}\) : 7
=> A = \(\dfrac{33}{119}\)
Ta có: \(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+...+\dfrac{68-65}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)=\dfrac{4}{3}.\dfrac{33}{68}=\dfrac{11}{17}\)
Tính:
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\frac{99}{202}\)
\(=\frac{66}{101}\)
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\Rightarrow A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{98.101}\right)\)
\(\Rightarrow A=4\left[\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{98}-\frac{1}{101}\right)\right]\)
\(\Rightarrow A=4\left[\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\right]\Rightarrow A=4\left(\frac{1}{3}.\frac{99}{202}\right)\Rightarrow A=4.\frac{33}{202}\)\(\Rightarrow A=\frac{66}{101}\)