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\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{999}.\dfrac{1}{1000}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\\ =1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
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Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
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Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.
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1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot\dfrac{0}{6}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot0\)
\(Q=0\)
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Bài 1: a) Ta có : \(\dfrac{-3}{x}=\dfrac{x}{-27}\Leftrightarrow\left(-3\right).\left(-27\right)=x.x\Leftrightarrow81=x^2\Leftrightarrow9^2=x^2\Leftrightarrow x=9\)
b) Do \(\dfrac{2}{3}\) của x là -150 nên x là: (-150) : \(\dfrac{2}{3}\) = -225
c) \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{18}\)
\(\Leftrightarrow x+2=18\)
\(\Leftrightarrow x=16\)
Bài 2:
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right).0\)
\(A=0\)
A = ( \(\dfrac{1}{2}\) + 1)(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)....(\(\dfrac{1}{999}\) + 1)
A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\).......\(\dfrac{1000}{999}\)
A = \(\dfrac{3.4.5......999}{3.4.5......999}\). \(\dfrac{1000}{2}\)
A = 1 \(\times\) 500
A = 500