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\(\begin{array}{l}\frac{7}{6}.3\frac{1}{4} + \frac{7}{6}.( - 0,25)\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 25}}{{100}}\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 1}}{4}\\ = \frac{7}{6}.[\left( {\frac{{13}}{4} + ( - \frac{1}{4})} \right)]\\ = \frac{7}{6}.\frac{{12}}{4}\\ = \frac{7}{6}.3\\ = \frac{7}{2}\end{array}\)
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
2 6/31 - 5/24 + 25/31 - 1 1/24 + 0,25
= 68/31 - 5/24 + 25/31 - 25/24 + 1/4
= (68/31 - 25/31) + (5/24 + 25/24 - 1/4)
= (68/31 - 25/31) + (5/24 + 25/24 - 6/24)
= 43/31 + 1
= 74/31
đây là toán lớp 7 á mik mới lớp 5 mik lm đc nek