Hướng dẫn tôi cách tính nhanh phân số sau
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
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WH
29 tháng 3 2018
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
MP
25 tháng 1 2016
A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
DD
5 tháng 1 2016
\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)
A.2=2/3.5+2/5.7+2/7.9+…+2/99.101
A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101
Vậy
A.2=1/3-1/101=98/303
A=98/303/2=49/303
Đúng không
4 tháng 1 2016
A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101
= 1/3 - 1/101 = 98/303
Vậy A = 98/303 : 2 = 49/303
NH
22 tháng 3 2019
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(A=\frac{1}{2}\left(\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(A=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
P/s: Có thể tính sai :<
22 tháng 3 2019
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
NM
24 tháng 3 2019
a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{14}{30}=\frac{7}{15}\)
LN
24 tháng 3 2019
a)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=2\left(1-\frac{1}{15}\right)\)
\(=2.\frac{14}{15}\)
\(=\frac{28}{15}\)
b)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)
\(...\)
23 tháng 1 2018
Ta có: \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(\Leftrightarrow A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\Rightarrow2A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{10}{11}:2=\frac{5}{11}\)
Vậy \(A=\frac{5}{11}\)
HB
23 tháng 1 2018
A = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
A = \(1-\frac{1}{11}\)
A = \(\frac{10}{11}\)
PM
23 tháng 4 2018
2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
=\(1-\frac{1}{15}=\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
23 tháng 4 2018
a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)
A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)
A=2(1-1/15)
A=1/2.14/15
A=7/15