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cau 1 : 0,43(34)=0,43+0,00(34)
=43/100+0,(34)/10^2
=43/100+34.0,(01)/10^2
=43/100+34.1/99.10^-2
=43/100+34/9900
=4291/9900
=> tử số là 4291
Câu 2 : Số 1,6(2) khi viết dưới dạng phân số tối giản sẽ có mẫu bằng 45
Câu 1 1/9=0.(1)
1/99=0,(01)
1/999=0,(001)
1/9999=0,(0001)
.......................
ta có
0,43(34)=0,43+0,00(34)
=43/100+0,(34)/10^2
=43/100+34.0,(01)/10^2
=43/100+34.1/99.10^-2
=43/100+34/9900
=4291/9900
=> tử số là 4291
Bài 1:
a) 0,24 = 6/25
b) 0,245 = 49/200
c) 2,5324 = 5/2
d) 0,5 = 1/2
a) \(0,\left(24\right)=\frac{24}{99}=\frac{8}{33}\)
b)\(0,2\left(45\right)=\frac{245-2}{990}=\frac{243}{990}=\frac{27}{110}\)
c)\(2,5\left(324\right)=2+0,5\left(324\right)=2+\frac{5324-5}{9990}=2+\frac{197}{370}=\frac{937}{370}\)
d) \(0,5\left(3\right)=\frac{53-5}{90}=\frac{48}{90}=\frac{8}{15}\)
Bài 2 : \(M=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{16-1}{90}}{\frac{5}{2}+\frac{5}{3}-\frac{83-8}{90}}\)
\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}=\frac{1}{5}\)
a ) -2/3 - x = 0,45 b) 1/3 - x( 7/12 +2) = 5/6
=> x = -2/3 -0,45 => 1/3 - x . 31/12 = 5/6
=> x = -67/60 => 31/12x = 1/3 - 5/6
vậy x = -67/60 => 31/12x = - 1/2
=> x = -1/2 : 31/12
=> x = -6/31 vậy x = -6/31
c) 2 2/3x + 8 2/3 = 2 1/3 d) 3/7 ( x+1) = 4/7
=> 8/3x + 26/3 = 7/3 => x+1 = 4/7 : 3/7
=> 8/3x = 7/3 -26/3 => x+1 = 4/3
=> 8/3x = -19/3 => x = 4/3 - 1
=> x = -19/3 : 8/3 = -19/8 => x = 1/3
vậy x= -19/8 vậy x = 1/3
e)1/3x + 2/5 ( x+1 ) =6 f) x/126 = -5/9 . 4/7
=>1/3x + 2/5x +2/5 = 6 => x/126 = -20/ 63
=> (1/3 + 2/5)x = 6 - 2/5 => x = -20/63 . 126
=>11/15x = 28/5 => x = -40
=> x= 28/5 : 11/15 vậy x= -40
=> x = 84/11
vậy x= 84/11
CHÚC BẠN HOK TỐT
a, \(\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
b, \(\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{2}{8}+\dfrac{1}{8}=\dfrac{3}{8}\)
c, \(\dfrac{-5}{12}+\dfrac{2}{-3}=\dfrac{-5}{12}+\dfrac{-8}{12}=\dfrac{-13}{12}\)
\(a,\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(b,\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{2}{8}+\dfrac{1}{8}=\dfrac{3}{8}\)
\(c,\dfrac{-5}{12}+\dfrac{2}{-3}=\dfrac{-5}{12}+\dfrac{-2}{3}=\dfrac{-5}{12}+\dfrac{-8}{12}=\dfrac{-13}{12}\)