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\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(17\frac{2}{31}=\frac{529}{31};6\frac{2}{31}=\frac{188}{31}\)
\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(=\frac{529}{31}-\left(\frac{15}{17}+\frac{188}{31}\right)\)
\(=\frac{529}{31}-\frac{3661}{527}\)
\(=\frac{172}{17}\)
\(\left(\frac{29}{31}-\frac{7}{8}\right)-\left(\frac{28}{31}-4\right)\)
\(=\frac{15}{248}-\left(-\frac{96}{31}\right)\)
\(=\frac{783}{248}\)
a, \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=31\frac{6}{16}+5\frac{9}{41}-36\frac{6}{13}\)
\(=\left(31\frac{6}{16}-31\frac{6}{16}\right)+5\frac{9}{41}\)
\(=0+5\frac{9}{41}=5\frac{9}{41}\)
b, \(\left(17\frac{29}{31}-3\frac{7}{8}\right)-\left(2\frac{28}{31}-4\right)=17\frac{9}{31}-3\frac{7}{8}-2\frac{28}{31}+4\)
(556/31-31/8)-(90/31-4)
3567/248-(90/31-4)
3567/248+34/31
3839/248
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot0\)
\(C=0\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Ta thấy 1/2-1/3-1/6=1/6-1/6=0
\(\Rightarrow C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x0\)
\(\Rightarrow C=0\)
Vậy...............
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
\(=15\frac{29}{31}-2\frac{7}{8}-2\frac{29}{31}+3=\left(15\frac{29}{31}-2\frac{29}{31}\right)-\left(2\frac{7}{8}+3\right)\)
\(=13+5\frac{7}{8}\)\(=18\frac{7}{8}\)