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a \(\frac{4}{18}+\frac{5}{13}+\frac{7}{9}+\frac{21}{13}\)
= \(\left(\frac{4}{18}+\frac{7}{9}\right)+\left(\frac{5}{13}+\frac{21}{13}\right)\)
= \(1+2\)
= \(3\)
b \(\frac{4}{3}-\frac{2}{7}-\frac{5}{7}+\frac{2}{3}\)
= \(\left(\frac{4}{3}-\frac{2}{3}\right)-\left(\frac{2}{7}+\frac{5}{7}\right)\)
= \(\frac{2}{3}-1\)
= \(-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(5\frac{3}{4}:3+2\frac{1}{4}.\frac{1}{3}-\frac{3}{8}=\frac{23}{4}:\frac{3}{1}+\frac{9}{4}.\frac{1}{3}=\frac{23}{12}+\frac{3}{4}=\frac{8}{3}\)
b ) \(\frac{3}{5}:\frac{5}{6}:\frac{6}{7}:\frac{7}{8}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{3.6.7.8}{5.5.6.7}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{24}{25}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{4049}{1400}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)có 9 p/số
\(=\frac{1}{10}.9=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}+\)\(\frac{1}{10}\)
\(=\frac{1}{10}.9\)
\(=\frac{9}{10}\)
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\(\frac{1}{10}+\frac{2}{20}+\frac{3}{30}+\frac{4}{40}+\frac{5}{50}+\frac{6}{60}+\frac{7}{70}+\frac{8}{80}+\frac{9}{90}\)
\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\)
\(=\frac{1}{10}\times9\)
\(=\frac{9}{10}\)
a)\(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{1}{3}\right)+3\)
\(=1+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3\)
\(=1+3+\frac{7}{9}\)
\(=4\frac{7}{9}\)
\(=\frac{43}{9}\)
b)\(\frac{8}{5}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{2}{5}-1\)
\(=\frac{4}{7}\cdot\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}\cdot2-1\)
\(=\frac{8}{7}-1\)
\(=\frac{1}{7}\)
a) \(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3=4\frac{7}{9}\)
b)\(\frac{8}{5}.\frac{4}{7}+\frac{4}{7}.\frac{2}{5}-1\)
\(=\frac{4}{7}.\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}.2-1\)
\(=\frac{8}{7}-\frac{7}{7}=\frac{1}{7}\)