Tính giá trị biểu thức:
A= \(\frac{4}{5.8}\)+ \(\frac{4}{8.11}\)+ ... + \(\frac{4}{305.308}\)
K
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7 tháng 8 2016
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\frac{99}{202}\)
\(=\frac{66}{101}\)
LH
7 tháng 8 2016
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
9 tháng 8 2015
S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48
S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )
S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )
S = 4/3 . ( 1/2 - 1/68 )
S = 4/3 . 33/68
S = 11/17
NT
2
10 tháng 5 2019
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{65.68}\)
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)
\(A=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{65}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\frac{33}{68}\)
\(A=\frac{11}{17}\)
~ Hok tốt ~
TN
10 tháng 5 2019
\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)
23 tháng 9 2020
a) \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot13}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}=\frac{1}{5}-\frac{1}{650}=\frac{129}{650}\)
b) \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)
sửa 199 -> 200
P/S : Lần sau đừng có đăng từng câu từng câu hỏi trên đây nhá
AF
23 tháng 9 2020
Bài giải
a, \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}\)
\(A=\frac{1}{5}-\frac{1}{650}=\frac{13}{650}-\frac{1}{650}=\frac{12}{650}=\frac{6}{325}\)
b, \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}\)
\(B=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)
11 tháng 3 2017
\(\frac{4}{2\cdot5}\)+\(\frac{4}{5\cdot8}\)+\(\frac{4}{8\cdot11}\)+.......+\(\frac{4}{8\cdot83}\)=\(\frac{4}{3}\) (\(\frac{3}{2\cdot5}\)+\(\frac{3}{5\cdot8}\) +......+\(\frac{3}{80\cdot83}\) )
=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{5}\) +\(\frac{1}{5}\) -\(\frac{1}{8}\) +..........+\(\frac{1}{80}\) -\(\frac{1}{83}\) )
=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{83}\) )
=\(\frac{4}{3}\)*\(\frac{81}{166}\)
=\(\frac{54}{83}\)
22 tháng 4 2018
Ta có: \(A=\frac{4}{5.8}+\frac{4}{8.11}+\frac{4}{11.14}+...+\frac{4}{305.308}\)
\(\frac{3}{4}A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{305.308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{303}{1540}\)
\(\Rightarrow A=\frac{303}{1450}\div\frac{3}{4}\)
\(\Rightarrow A=\frac{101}{385}\)
Nhớ k mình nha.
22 tháng 4 2018
\(A=\frac{4}{5.8}+\frac{4}{8.11}+\frac{4}{11.14}+...+\frac{4}{305.308}\)
\(\frac{3}{4}A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{305.308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8} -\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{308}=\frac{303}{1540}\)
\(A=\frac{308}{1540}:\frac{3}{4}=\frac{4}{15}\)
A = 4/5.8 + 4/8.11 + ... + 4/305.308
A = 4. ( 1/5.8 + 1/8.11 + ... + 1/305 . 108 )
A = 4 x 1/3 ( 3/5.8 + 3/8.11 + ... + 3/105.108 )
A = 4/3 ( 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/105 - 1/108 )
A = 4/3 ( 1/5 - 1/108 )
A = 4/3 . 103/540
A = 617/540
Vậy A = 617/540 !!
\(A=\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+\frac{4}{11\cdot14}+...+\frac{4}{305\cdot308}\)
\(\frac{3}{4}A=\frac{3}{4}\left(\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+...+\frac{4}{305\cdot308}\right)\)
\(\frac{3}{4}A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{305\cdot308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{303}{1540}\Rightarrow A=\frac{303}{1540}:\frac{3}{4}=\frac{101}{385}\)