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Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10
A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ... + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)
A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]
A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2)
A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2
A = 1/2 × 5 = 5/2
Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10
A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ... + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)
A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]
A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2)
A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2
A = 1/2 × 5 = 5/2
a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
\(=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\) ( có 9 thừa số )
\(=-\left(\frac{3}{4}\right)\cdot\left(\frac{8}{9}\right)\cdot\left(\frac{15}{16}\right)\cdot...\left(\frac{99}{100}\right)\) ( có 9 thừa số nên tích sẽ âm )
\(=-\left(\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7\cdot6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10}\right)\)
\(=-\left(\frac{11}{20}\right)\)
Bài giải
\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot...\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)
\(=-\left(\frac{3\cdot8\cdot...\cdot99}{4\cdot9\cdot...\cdot100}\right)=-\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7\cdot6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10}=-\frac{11}{20}\)