\(\left(2,8x-32\right):\dfrac{2}{3}\)=-90
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a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\Rightarrow\left(2,8x-32\right)=-90\cdot\frac{2}{3}\)
\(\Rightarrow\left(2,8x-32\right)=-60\)
\(\Rightarrow2,8x=-60+32\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
\(\left(2,8x-32\right)\div\frac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
\(\left(4,5-2x\right)\times1\frac{4}{7}=\frac{11}{14}\)
\(\Rightarrow4,5-2x=\frac{1}{2}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Bài 1 :
Để phân số \(A=\dfrac{n+6}{n-1}\in Z\left(n\in N\right)\) thì :
\(n+6⋮n-1\)
Mà \(n-1⋮n-1\)
\(\Leftrightarrow7⋮n-1\)
Vì \(n\in N\Leftrightarrow n-1\in N;n-1\inƯ\left(5\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}n-1=1\Leftrightarrow n=2\\n-1=5\Leftrightarrow n=6\end{matrix}\right.\) \(\left(tm\right)\)
Vậy ...............
Bài 2 :
Ta có :
\(11n+7⋮n\)
Mà \(n⋮n\)
\(\Leftrightarrow\left\{{}\begin{matrix}11n+7⋮n\\11n⋮n\end{matrix}\right.\)
\(\Leftrightarrow7⋮n\)
Vì \(n\in N\Leftrightarrow n\inƯ\left(7\right)=\left\{1;7\right\}\)
Vậy ................
bài 3 :
a) \(\left(5+\dfrac{4}{7}\right):x=13\)
\(\dfrac{39}{7}:x=13\)
\(x=\dfrac{39}{7}:13\)
\(x=\dfrac{1}{7}\)
Vậy .................
b) \(\left(2,8x+32\right):\dfrac{2}{3}=90\)
\(2,8x+32=90.\dfrac{2}{3}\)
\(2,8x+32=60\)
\(2,8x=60-32\)
\(2,8x=28\)
\(x=28:2,8\)
\(x=10\)
Vậy .........
( 2,8x - 32 ) : 2/3 = -90
( 2,8x - 32 ) = -60
2,8x = -60 +32
2,8x = -28
x = -28 : 2,8
x = -10
(2,8x - 32) : 2\3 = -90
2,8x - 32 = -90 . 2/3
2,8x - 32 = -60
2,8x = -60 + 32
2,8x = -28
x = -28 : 2,8
x = -10
\(\left(2,8x-32\right):\frac{2}{3}=-90\\ \frac{14}{5}x-32=\left(-90\right)\cdot\frac{2}{3}\\ \frac{14}{5}x-32=-60\\ \frac{14}{5}x=\left(-60\right)+32\\ \frac{14}{5}x=-28\\ x=\left(-28\right):\frac{14}{5}\\ x=\left(-28\right)\cdot\frac{5}{14}\\ x=-10\)Vậy x = -10
\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90.\frac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8=-10\)
Vậy \(x=-10\)
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)
\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)
\(\Leftrightarrow2,8x=-60+32=-28\)
\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)
d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)
Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.
mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha
2 , 8 x - 32 = -90.\(\dfrac{2}{3}\)
2 , 8 x - 32 = -60
2 , 8 x = -60 + 32
2 , 8 x = -28
x = -28 : 2, 8
x = -10
\(\left(2,8x-32\right):\dfrac{2}{3}=-90\\ \left(2,8x-32\right)=-90.\dfrac{2}{3}=-60\\ 2,8x=-60+32=-28\\ x=-28:2,8=-10\)