\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}\)​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

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)353/180

b)-79/225

$-\frac{79}{225}$ **** giùm mình đi

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.211,0465116

Sai thôi nha

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!

\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)

\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)

\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)

\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)

\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)

\(A=1,3\left(2\right)\)

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

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

\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\frac{75}{53}:\frac{50}{53}\)

\(=\frac{13}{30}+\frac{7}{3}\)

\(=\frac{83}{30}\)

Bài 2:

a) \(\frac{4}{9}+x=\frac{-5}{3}\)

\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)

\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)

Vậy: \(x=\frac{-19}{9}\)

b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)

\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)

\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)

c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)

\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)

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