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\(\Rightarrow\frac{11}{12}+\frac{11}{12}-\frac{11}{23}+...+\frac{11}{89}-\frac{11}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{11}{150}+x=\frac{2}{3}\)
=>\(x=\frac{2}{3}-\frac{11}{150}\)
=>x=\(\frac{89}{150}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{99}{100}+x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{99}{100}=\dfrac{-97}{300}\)
Vậy \(x=\dfrac{-97}{300}\)
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\(\left(\frac{11}{1.12}+\frac{11}{12.23}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\Leftrightarrow\frac{99}{100}+x=\frac{5}{3}\Leftrightarrow x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow x=\frac{297}{300}=\frac{ }{100}\)
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