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D=\(-\dfrac{1}{4.5}\)+(\(-\dfrac{1}{5.6}\))+(\(-\dfrac{1}{6.7}\))+(\(-\dfrac{1}{7.8}\))+(\(-\dfrac{1}{8.9}\))+(\(-\dfrac{1}{9.10}\))
D=\(-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
D=\(-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
D=\(-\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)
D=\(-\dfrac{3}{20}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{100}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)...\left(\frac{100}{100}+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}...\frac{101}{100}\)
\(=\frac{3.4...101}{2.3...100}\)
\(=\frac{101}{2}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{100}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)...\left(\frac{100}{100}+\frac{1}{100}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{101}{100}\)
\(=\frac{101}{2}\)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+1/9.10
=1/2(1-1/9)+1/90
=1/2.8/9+1/90=4/9+1/90=41/90
\( S = 1-1/5 +1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21+1/21-1/25+1/25-1/29. \)
\(S= 1- 1/29 \)
\(S=\frac{28}{29}\)
Nếu mình ko nhầm!
F = \(\frac{1}{2}\) . \(\frac{2}{3}\) ..... \(\frac{98}{99}\) .\(\frac{99}{100}\)
\(\Leftrightarrow\)F = \(\frac{1.2.3...98.99}{2.3.4...99.100}\)
\(\Leftrightarrow\)F = \(\frac{1}{100}\)
Vậy F =\(\frac{1}{100}\)
\(F=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)
\(F=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{99}{100}\right)\)
F có : ( 99 - 1 ) : 1 + 1 = 99 phân số
=> F mang dấu âm
=> \(F=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{99}{100}\right)\)
=> \(F=-\left(\frac{1\cdot2\cdot...\cdot99}{2\cdot3\cdot...\cdot100}\right)\)
=> \(F=-\frac{1}{100}\)
Sửa đề : \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{67.68}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{68}\)
\(=\frac{1}{4}-\frac{1}{68}=\frac{4}{17}\)
10.Vì 1 bàn tay cộng một bàn tay bằng 10>hehe