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Bài 1:
a) \(0,\left(3\right)+3\dfrac{1}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{19}{45}\)
\(=\dfrac{184}{45}\)
b) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{1789}{990}\)
Bài 2:
a) \(0,\left(37\right)x=1\)
\(\Leftrightarrow\dfrac{37}{99}.x=1\)
\(\Leftrightarrow x=1:\dfrac{37}{99}\)
\(\Leftrightarrow x=\dfrac{99}{37}\)
b) \(0,\left(26\right)x=1,2\left(31\right)\)
\(\Leftrightarrow\dfrac{26}{99}x=\dfrac{1219}{990}\)
\(\Leftrightarrow x=\dfrac{1219}{990}:\dfrac{26}{99}\)
\(\Leftrightarrow x=\dfrac{1219}{260}\)
Chúc bạn học tốt!
a)ta có: 0, (37) + 0, (62) = 1
\(\Rightarrow\)\(\dfrac{37}{99}+\dfrac{62}{99}=1\left(ĐPCM\right)\)
b)ta có: 0, (33).3=1
\(\Rightarrow\)\(\dfrac{1}{3}.3=1\left(ĐPCM\right)\)
a) Ta có:
0, (37) = 0, (01) . 37 = \(\dfrac{1}{99}\) . 37 = \(\dfrac{37}{99}\)
0, (62) = 0, (01) . 62 = \(\dfrac{1}{99}\) . 62 = \(\dfrac{62}{99}\)
\(\Rightarrow\)0, (37) + 0, (62) = \(\dfrac{37}{99}\) + \(\dfrac{62}{99}\) = \(\dfrac{99}{99}\)= 1
Vậy 0, (37) + 0, (62) = 1 (ĐPCM)
b) Ta có:
0, (33) = 0, (01) . 33 = \(\dfrac{1}{99}\) . 33 = \(\dfrac{33}{99}\)
\(\Rightarrow\)0, (33) . 3 = \(\dfrac{33}{99}\) . 3 =\(\dfrac{99}{99}\) = 1
Vậy 0, (33) . 3 = 1 (ĐPCM)
tick mk nhé
Bài 1:
a) \(0,\left(3\right)+3\frac{1}{3}+0,\left(31\right)\)
\(=\frac{1}{3}+\frac{10}{3}+\frac{31}{99}\)
\(=\frac{11}{3}+\frac{31}{99}\)
\(=\frac{394}{99}.\)
b) \(\frac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\frac{4}{9}+\frac{1219}{990}-\frac{13}{99}\)
\(=\frac{553}{330}-\frac{13}{99}\)
\(=\frac{139}{90}.\)
Bài 2:
\(0,\left(37\right).x=1\)
\(\Rightarrow\frac{37}{99}.x=1\)
\(\Rightarrow x=1:\frac{37}{99}\)
\(\Rightarrow x=\frac{99}{37}\)
Vậy \(x=\frac{99}{37}.\)
Chúc bạn học tốt!
Phương Nguyễn Mai Bạn thử xem ở đây nhé:
Lý thuyết số thập phân hữu hạn. số thập phân vô hạn tuần ...
1) b) \(0,555..=0,\left(5\right)=\frac{5}{9}\)
a) 0,555=\(\frac{555}{1000}=\frac{111}{200}\)
c) \(0,25454..=0,2\left(54\right)=\frac{14}{55}\)
2) a) \(1,\left(6\right).2,\left(3\right):0,\left(7\right)=\frac{5}{3}.\frac{7}{3}:\frac{7}{9}=\frac{35}{9}:\frac{7}{9}=5\)
b) \(0,\left(37\right)+0,\left(62\right)=\frac{37}{99}+\frac{62}{99}=1\)
c) \(0,\left(33\right).3=\frac{1}{3}.3=1\)
a) 0,(37)+0,(62) = 1
Có 0.(37)=\(\frac{37}{99}\)và 0.(62) = \(\frac{62}{99}\)
\(\frac{37}{99}\)+ \(\frac{62}{99}\)= 1
\(\Rightarrow0,\left(37\right)+0.\left(62\right)=1\)
b)\(0,\left(37\right)\times3=1\)
Có: \(0,\left(37\right)=\frac{37}{99}\)
\(\frac{37}{99}\times3=1\)
\(\Rightarrow0\left(37\right)\times3=1\)
Ta có: \(0,\left(3\right)+\frac{31}{3}+0,4\left(2\right)=\frac{3}{9}+\frac{31}{3}+\frac{42-4}{90}=\frac{1}{3}+\frac{31}{3}+\frac{19}{45}=\frac{32}{3}+\frac{19}{45}=\frac{499}{45}.\)
\(\frac{4}{9}+0,\left(13\right)=\frac{4}{9}+\frac{13}{99}=\frac{44}{99}+\frac{13}{99}=\frac{57}{99}=\frac{19}{33}\)
\(0,\left(37\right).x\Rightarrow\frac{37}{99}.x=1\)
\(\Rightarrow x=1:\frac{37}{99}=\frac{99}{37}\)
\(0,\left(26\right).x=1,2\left(31\right)\)
\(\Rightarrow\frac{26}{99}.x=\frac{1219}{990}\)
\(\Rightarrow x=\frac{1219}{990}:\frac{26}{99}=\frac{1219}{260}\)
\(0,\left(37\right)+0,\left(62\right)=0,\left(99\right)\)
Theo quy ước làm tròn số ta dược :
\(0,\left(99\right)\approx1\) (đpcm)
b) Làm tương tự câu a) ta có :
\(0,\left(33\right).3=0,\left(99\right)\approx1\) (đpcm)
\(0,\left(37\right)+0,\left(62\right)=\frac{37}{99}+\frac{62}{99}=\frac{99}{99}=1\)
\(0,\left(33\right).3=\frac{33}{99}.3=\frac{1}{3}.3=\frac{3}{3}=1\)
Ta có:
Chọn đáp án D