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[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)
\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)
Tính nốt vì số dữ quá , lần sau để số ít thôi
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)
Ta có:
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{1}{3}+\frac{1}{2}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).\left(\frac{5}{6}-\frac{5}{6}\right)\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1100}\right).0\)
\(=0\)