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\(P=\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}=\left(\frac{4}{11}+\frac{3}{22}\right)+\left(\frac{5}{2}+\frac{1}{30}+\frac{13}{60}\right)=\frac{1}{2}+\frac{11}{4}=\frac{13}{4}\)
\(Q=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}=\frac{1.2.3....19}{2.3.4....20}=\frac{1}{20}\)
Thực hiện phép tính
a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
= \(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)
= \(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)
= \(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)
= \(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)
= \(\frac{-7}{20}+\frac{1}{2}\)
= \(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)
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\(\left(\frac{12}{32}+\frac{5}{-20}-\frac{10}{24}\right):\frac{2}{3}=\left(\frac{1}{8}-\frac{10}{24}\right):\frac{2}{3}=-\frac{7}{24}:\frac{2}{3}=-\frac{7}{16}\)
\(4\frac{1}{2}:\left(2,5-3\frac{3}{4}\right)+\left(\frac{1}{2}\right)^2=\frac{9}{2}:\left(2,5-\frac{15}{4}\right)+\frac{1}{4}=\frac{9}{2}:-\frac{5}{4}+\frac{1}{4}=-\frac{18}{5}+\frac{1}{4}=-\frac{67}{20}\)
a)\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)
b)\(\frac12+\frac34-\left(\frac34-\frac45\right)\)
=\(\frac12+\frac34-\frac34+\frac45\)
=\(\left(\frac12+\frac45\right)+\left(\frac34-\frac34\right)\)
=\(\left(\frac{5}{10}+\frac{8}{10}\right)+0\)
=\(\frac{13}{10}\)
\(\left(1-\frac{4}{1}\right).\left(1-\frac{4}{9}\right)...\left(1-\frac{4}{2017^2}\right)\)
\(=\left(\frac{1-4}{1}\right).\left(\frac{9-4}{9}\right)...\left(\frac{2017^2-4}{2017^2}\right)\)
\(=\left(-3\right).\left(\frac{1.5}{3.3}\right).\left(\frac{3.7}{5.5}\right)...\left(\frac{2015.2019}{2017.2017}\right)\)
\(=-\left(3\right).\frac{1}{3}.\frac{2019}{2017}=-\frac{2019}{2017}\)






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