Question

Tính

a,\(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)

b,\(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)

c, \(3\frac{1}{2}.\frac{4}{49}-\left[2,\left(4\right).2\frac{5}{11}\right]:\frac{-42}{53}\)

Answer 1

a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)

\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)

\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)

b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)

\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)

\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)

c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)

\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)

\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)