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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) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
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 = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)
Ta có:
\(\left(\right. a - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (1)
Và: \(a + 1 = b + 2 = c + 3\)
\(\Rightarrow a = b + 2 - 1 = b + 1\)
Thay vào (1) ta có:
\(\left(\right. b + 1 - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (2)
Mà: \(b + 2 = c + 3\)
\(\Rightarrow c = b + 2 - 3 = b - 1\)
Thay vào (2) ta có:
\(\left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 1 - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 4 \left.\right) = 0\)
\(\Rightarrow \left[\right. b = - \frac{2}{3} \\ b = - \frac{1}{2} \\ b = 4\)
TH1 khi b=\(- \frac{2}{3}\)
\(\Rightarrow a = b + 1 = - \frac{2}{3} + 1 = \frac{1}{3}\)
\(\Rightarrow c = b - 1 = - \frac{2}{3} - 1 = - \frac{5}{3}\)
TH2 khi \(b = - \frac{1}{2}\)
\(\Rightarrow a = b + 1 = - \frac{1}{2} + 1 = \frac{1}{2}\)
\(\Rightarrow c = b - 1 = - \frac{1}{2} - 1 = - \frac{3}{2}\)
TH3 khi \(b = 4\)
\(\Rightarrow a = b + 1 = 4 + 1 = 5\)
\(\Rightarrow c = b - 1 = 4 - 1 = 3\)
sai mình xin lỗi
Câu 1:
c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)
\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)
\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)
=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
=>x=10
Câu 2:
d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)
\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)
\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)
\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)
\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)
\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)
\(A=1,3\left(2\right)\)
Tính:
0,4(3) + 0,6(2) . \(2\frac{1}{2}\).[(\(\frac{1}{2}+\frac{1}{3}\)) : 0,5(8)] : \(\frac{50}{53}\)
0,4(3) + 0,6(2). \(2\frac{1}{2}\).\(\left[\left(\frac{1}{2}+\frac{1}{3}\right):0,5\left(8\right)\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\frac{75}{53}:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{7}{3}\)
\(=\frac{83}{30}\)
2.09
Ta có: \(0,4\left(3\right)+0,6\left(2\right)\cdot2\frac12-\frac{\frac12+\frac13}{0,5\left(8\right)}:\frac{50}{53}\)
\(=\frac{13}{30}+\left\lbrack0,6+0,0\left(2\right)\right\rbrack\cdot\frac52-\frac{\frac56}{0,5+0,0\left(8\right)}\cdot\frac{53}{50}\)
\(=\frac{13}{30}+\left(\frac35+\frac{1}{45}\right)\cdot\frac52-\frac56:\left(\frac12+\frac{4}{45}\right)\cdot\frac{53}{50}\)
\(=\frac{13}{30}+\frac{28}{45}\cdot\frac52-\frac56:\frac{53}{90}\cdot\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{15}-\frac56\cdot\frac{90}{53}\cdot\frac{53}{50}=\frac{13}{30}+\frac{28}{30}-\frac56\cdot\frac{90}{50}\)
\(=\frac{41}{30}-\frac{15}{10}=\frac{41}{30}-\frac32=\frac{41}{30}-\frac{45}{30}=\frac{-4}{30}=-\frac{2}{15}\)