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\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).....\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}\)
\(=\frac{101}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\times\left(1-\dfrac{1}{6}\right)\times\dots\times\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\) (sửa đề)
\(=\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dots\times\dfrac{98}{99}\times\dfrac{99}{100}\)
\(=\dfrac{2\times3\times4\times5\times\dots\times98\times99}{3\times4\times5\times6\times\dots\times99\times100}\)
\(=\dfrac{2}{100}\)
\(=\dfrac{1}{50}\)
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A=(2-3+4-5) +(6-7+8-9)+.......=(96-97+98-99)+100
A=0+0+0+.....+0+100
A=100
BÀI D EM NGẠI VIẾT
a) \(A=2+1+1+...+1=2+49=51.\)
b) \(B=1,7+1,7+...+1,7=1,7.10=17.\)
c) \(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow D=1-\frac{1}{10}=\frac{9}{10}.\)
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\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
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