Question

[KurokoTetsuya]

Tính:

\(\left\{\left[\left(6,2:0,31-\dfrac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}\right]\)

b) \(0,4\left(3\right)+0,6\left(2\right).2\dfrac{1}{2}.\left[\left(\dfrac{1}{2}+\dfrac{1}{3}\right):0,5\left(8\right)\right]:\dfrac{50}{53}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)

\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)

\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)

b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)

\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)