![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left[0,\left(37\right)+0,\left(62\right)\right]\cdot x=10\)
=> \(\left[\frac{37}{99}+\frac{62}{99}\right]\cdot x=10\)
=> \(1\cdot x=10\Rightarrow x=10\)
b) \(\frac{0,\left(12\right)}{1,\left(6\right)}=\frac{\frac{12}{99}}{\frac{5}{3}}=\frac{12}{99}\cdot\frac{3}{5}=\frac{4}{55}\)
=> \(\frac{4}{55}=x:0,\left(4\right)\)
=> \(\frac{4}{55}=x:\frac{4}{9}\)
=> \(x:\frac{4}{9}=\frac{4}{55}\)
=> \(x=\frac{4}{55}\cdot\frac{4}{9}=\frac{16}{495}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 0,(37)+0,(62)=1
ta có : 0,(37)=37/99
0,(62)=62/99
=> 0,(37)+0,(62)=37/99+62/99=99/99=1
Vậy 0,(37)+0,(62)=1
b, 0,(33).3=1
ta có : 0,(33)=33/99=1/3
=> 0,(33).3=1/3.3=1
Vậy 0,(33).3=1
0,(37)+0,(62)=0,(99)
Theo quy ước làm tròn số ta dược :
0,\left(99\right)\approx10,(99)≈1 (đpcm)
b) Làm tương tự câu a) ta có :
0,\left(33\right).3=0,\left(99\right)\approx10,(33).3=0,(99)≈1 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=5/9.2,(37)+0,(5).3,(62)-2 căn bậc hai (-2/3)^2
B=5/9.235/99+5/9.359/99-4/3
B=5/9.(235/99+359/99)-4/3
B=5/9.6-4/3
B=10/3-4/3
B=2
K MK NHA
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{-31}{37}\right)^5.\left(\frac{37}{62}\right)^5\)
\(=\left(\frac{-31}{37}.\frac{37}{62}\right)^5\)
\(=\left(\frac{-1}{2}\right)^5\)
\(=\frac{-1}{32}\)
\(\left(\frac{-31}{37}\right)^5.\left(\frac{37}{62}\right)^5\)
\(=\left(\frac{-31}{37}.\frac{37}{62}\right)^5\)
\(=\left(\frac{-1}{2}\right)^5\)
\(=\frac{-1}{32}\)
học tốt
0, ( 37 ) + 0, ( 62 )
= 0 , ( 99 )
\(\approx\)1
Hk tốt
0,(37) + 0,(62) =
\(\frac{37}{99}+\frac{62}{99}=\frac{99}{99}\)
\(\frac{99}{99}=1\)