Question

Tính hợp lý nếu có thể:

a) A =  0,4(3)+0,6(2).\(2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:50\frac{50}{53}\)

b) B =  [0,(5).0,(2)]:\(\left(3\frac{1}{3}:\frac{33}{25}\right)-\left(\frac{2}{5}.1\frac{1}{3}\right):\frac{4}{3}\)

Answer 1

a) 0,4(3) = \(\frac{4,\left(3\right)}{10}=\frac{4+\frac{1}{3}}{10}=\frac{13}{30}\); 0,6(2) = \(\frac{6,\left(2\right)}{10}=\frac{6+\frac{2}{9}}{10}=\frac{56}{90}=\frac{28}{45}\); 0,5(8) = \(\frac{5,\left(8\right)}{10}=\frac{5+\frac{8}{9}}{10}=\frac{53}{90}\)

Vậy A  = \(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{53}{90}}:\frac{2700}{53}\) = \(\frac{13}{30}+\frac{14}{9}-\frac{5}{6}.\frac{90}{53}.\frac{53}{2700}=\frac{13}{30}+\frac{14}{9}-\frac{1}{36}=\frac{353}{180}\)

b) 0,(5) = 5/9; 0,(2) = 2/9

B = \(\left(\frac{5}{9}.\frac{2}{9}\right):\left(\frac{10}{3}.\frac{25}{33}\right)-\left(\frac{2}{5}.\frac{4}{3}\right):\frac{4}{3}\)

B = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)

Answer 2

a)353/180

b)-79/225

Answer 3

$-\frac{79}{225}$ **** giùm mình đi

Answer 4

79/225 . Ấn **** cho tớ nhé.

Answer 5

$-/frac{79}{225}$

 

Answer 6

tại sao 0,(5)=5/9