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a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)
A =\(1-\frac{1}{2^7}\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=1-\frac{1}{11}\)
\(A=\frac{10}{11}\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)
\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\left(2\right)\)
Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Rightarrow B=1-\frac{1}{2^7}\)
\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)
HOK TOT
1/6 + 1/12 + 1/20 + 1/30 + 1/42
= 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
= 1/2 - 1/7
= 5/14
Chúc bạn may mắn......mình chính là Đào Minh Tiến!
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{72}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{9}{18}-\frac{2}{18}=\frac{7}{18}\)
dễ mà phân tích các mẫu ra là các tích của 2 số gần liên tiếp rồi áp dụng phân số ai cập thui
A=1/2+1/6+1/12+1/20+1/30
A=1/1x2+1/2x3+1/3x4+1/3x5+1/5x6
A=1/1-1/2+1/2-1/3+.......+1/5-1/6
A=1/1-1/6
A=5/6
a) Ta có\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\)
b) Ta có \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2048}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)(1)
Đặt S = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
Lấy 2S trừ S ta có :
2S - S \(=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\right)\)
\(S=1-\frac{1}{2048}\)
Khi đó (1) <=> \(1-\left(1-\frac{1}{2048}\right)=1-1+\frac{1}{2048}=\frac{1}{2048}\)
\(a,\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{90}+\frac{2}{110}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{90}+\frac{1}{110}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=1-\frac{2}{11}\)
\(=\frac{9}{11}\)
Đặt phép tính trên là \(A\)
Có: \(A=\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+....+\frac{1}{90}\right)-x=\frac{19}{24}\)
\(A=\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{9.10}\right)-x=\frac{19}{24}\)
\(A=\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)-x=\frac{19}{24}\)
\(A=\left(\frac{1}{3}-\frac{1}{9}\right)-x=\frac{19}{24}\)
\(A=\frac{2}{9}-x=\frac{19}{24}\)
\(x=\frac{2}{9}-\frac{19}{24}=-\frac{41}{72}\)
\(\Rightarrow x=-\frac{41}{72}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9+\left(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+...+\frac{1}{9\cdot10}\right)\)
\(A=9+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9+\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Đặt\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}...+\frac{1}{110}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{10.11}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
\(S=1-\frac{1}{11}\)
\(S=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)