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\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}\)đó Michiel Girl Mít ướt
Đó là công thức đưa 1 số thập phân vô hạn tuần hoàn sang phân số đó Michiel Girl Mít ướt
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
a) \(-5,13:\left(5\frac{5}{28}-1\frac{8}{9}.1,25+1\frac{16}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{17}{9}.1,25+\frac{79}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=-5,13:\frac{57}{14}\)
\(=-\frac{63}{50}\)
b) \(\left(3\frac{1}{3}.1,9+19,5:4\frac{1}{3}.\left(\frac{62}{75}-\frac{4}{25}\right)\right)\)
\(=\left(\frac{10}{3}.1,9+19,5:\frac{13}{3}\right).\left(\frac{62}{75}-\frac{4}{25}\right)\)
\(=\left(\frac{19}{3}+\frac{9}{2}\right).\frac{2}{3}\)
\(=\frac{65}{6}.\frac{2}{3}\)
\(=\frac{65}{9}\)
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Ta có:
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}.\)
\(0,6\left(2\right)=\frac{62-6}{90}=\frac{56}{90}=\frac{28}{45}.\)
\(0,6\left(8\right)=\frac{68-6}{90}=\frac{62}{90}=\frac{31}{45}.\)
Vậy:
\(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{31}{45}}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{75}{62}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{159}{124}\)
\(=\frac{179}{90}-\frac{159}{124}\)
\(=\frac{3943}{5580}.\)
Chúc bạn học tốt!
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 = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)
\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)
\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)
\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)
\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)
\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)
\(A=1,3\left(2\right)\)