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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=\(-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}.+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{10}\right)\)=\(-1.\frac{3}{20}=\frac{-3}{20}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{72}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{9}{18}-\frac{2}{18}=\frac{7}{18}\)
dễ mà phân tích các mẫu ra là các tích của 2 số gần liên tiếp rồi áp dụng phân số ai cập thui
\(A=10.\left(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{71}{72}+\frac{89}{90}\right)\)
Đặt \(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{71}{72}+\frac{89}{90}\)
\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(B=1+1+1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8,1\)
Ta có \(A=10.B=10.B=10.8,1=81\)
Vậy \(A=81\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9+\left(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+...+\frac{1}{9\cdot10}\right)\)
\(A=9+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9+\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
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
A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
A = \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)+ \(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)
A = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)\(+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(+\frac{1}{9}-\frac{1}{10}\)A = \(\frac{1}{5}-\frac{1}{10}\)
A = \(\frac{1}{10}\)
ta có :1/30+1/42+1/56+1/72+1/90=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10=1/5-1/6+1/6-1/7=1/7-1/8+1/8-1/9+1/9-1/10=1/5-1/10=1/10