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\(2\frac{4}{3}+\frac{4}{12}=\frac{2.3+4}{3}+\frac{4}{12}=\frac{10}{3}+\frac{4}{12}=\frac{40}{12}+\frac{4}{12}=\frac{44}{12}=\frac{11}{3}\)
\(2\frac{4}{3}=\frac{10}{3}\)
\(\frac{10}{3}+\frac{4}{12}=\frac{120}{36}+\frac{12}{36}\)
\(=\frac{132}{36}=\frac{22}{6}=\frac{11}{3}\)
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2\(\frac{2}{12}\)+ X = \(\frac{5}{7}\)+ \(\frac{8}{12}\)
\(\frac{13}{6}\)+ X=\(\frac{29}{21}\)
X= \(\frac{29}{21}\)- \(\frac{13}{6}\)
X=\(\frac{-11}{14}\)
\(|2^2_{12}\)\(+x=\frac{5}{7}+\frac{8}{12}\)
\(2+x=\frac{5}{7}+\frac{1}{2}\)
\(2+x=\frac{17}{14}\)
\(x=\frac{17}{14}-2\)
\(x=\frac{-11}{14}\)
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\(đổi9\frac{3}{2}=\frac{21}{2}\)
\(\frac{9}{2}+\frac{21}{2}=\frac{30}{2}=15\)
\(=15\)
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\(5-1,5-1\frac{1}{2}\)
=\(5-1,5-1,5\)
=\(5-\left(1,5+1,5\right)\)
=\(5-3\)
=\(2\)
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Ta có: \(\frac{54}{195}-\frac{1}{2}=\frac{108}{390}-\frac{195}{390}\)
\(=-\frac{87}{390}=-\frac{29}{130}\)
Vậy \(\frac{54}{195}-\frac{1}{2}=-\frac{29}{130}\)
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Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(12\frac{4}{5}\)=\(\frac{64}{5}\).
Ta có :
\(\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(12\frac{4}{5}+\frac{4}{5}=\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(2+4\frac{2}{6}=2+\frac{26}{6}=\frac{12}{6}+\frac{26}{6}\)\(=\frac{38}{6}=\frac{19}{3}\)
=38/6
k nha