\(A=\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}\\ =\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{31}{11}\\ =\left(\dfrac{30}{6}+\dfrac{15}{6}\right):\left(\dfrac{95}{30}-\dfrac{126}{30}\right)-\dfrac{31}{11}\\ =\dfrac{45}{6}:\dfrac{-21}{30}-\dfrac{31}{11}\\ =\dfrac{15}{2}\times\dfrac{-10}{7}-\dfrac{31}{11}=-\dfrac{75}{7}-\dfrac{31}{11}=-\dfrac{825}{77}-\dfrac{217}{77}=\dfrac{-1042}{77}\)

\(B=\left(-6\right).10:\left[-0,25+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{-1}{4}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{2}\right)+1\dfrac{3}{4}=120+1\dfrac{3}{4}=121\dfrac{3}{4}\)

\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)

= \(\left(\frac{10}{3}+\frac{5}{2}\right):\left(\frac{19}{6}-\frac{21}{5}\right)-\frac{11}{31}\)

= \(\frac{35}{6}:\frac{-31}{30}-\frac{11}{31}\)

= \(\frac{175}{-31}\)- \(\frac{11}{31}\)

= -6

Nếu mình đúng thì các bạn k mình nhé

là bằng -6 nhé

mk mới chỉ học lớp 5 thôi

\(\left(3\frac{1}{3}+2.5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)

\(=\left(\frac{10}{3}+\frac{5}{2}\right):\left(\frac{19}{6}-\frac{21}{5}\right)-\frac{1}{3}\)

\(=\left(\frac{20}{6}+\frac{15}{6}\right):\left(\frac{95}{30}-\frac{126}{30}\right)-\frac{1}{3}\)

\(=\frac{35}{6}:\frac{-31}{30}-\frac{1}{3}\)

\(=-\frac{175}{31}-\frac{1}{3}\)

\(=-\frac{525}{93}-\frac{31}{93}\)

\(=-\frac{556}{93}\)

A \(=\frac{35}{6}.\left(\frac{-30}{31}\right)-\frac{11}{31}=\frac{-175}{31}-\frac{11}{31}=\frac{-186}{31}=-6\)

\(B=-60:\left(\frac{-1}{2}\right)+\frac{7}{4}=30+\frac{7}{4}=\frac{127}{4}\)

=(10/3+5/2):(19/6-21/5)-11/31

=35/6:(-31/30)-11/31

=-175/31-11/31

=-6

CHÚC BN HOK TỐT NHA!! ^^

[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)

\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)