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\(\left(2,8.x-32\right):\frac{2}{3}=90\)
\(2,8.x-32=90.\frac{2}{3}\)
\(2,8.x-32=60\)
\(2,8.x=60+32\)
\(2,8.x=92\)
\(x=92:2,8\)
\(x=\frac{230}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2,8:x-32\right):\frac{2}{3}=-90\)
\(2,8:x-32=\left(-90\right).\frac{2}{3}\)
\(2,8:x-32=-60\)
\(2,8:x=-28\)
\(x=\left(-28\right):2,8\)
\(x=-10\)
\(4-\left|x-2012\right|=\left(-2\right)^2\)
\(4-\left|x-2012\right|=-4\)
\(\left|x-2012\right|=8\)
\(\Rightarrow\orbr{\begin{cases}\left|x-2012\right|=8\\\left|x-2012\right|=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=2020\\x=2004\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2,8 . x - 32) : 2/3 = -90
(2,8 . x - 32) = -90 x 2/3
(2,8 . x - 32) = -60
2,8 . x = -60 + 32
2,8 . x = -28
x = -28 : 2,8
x = -10
`(2,8 . x - 32) : 2/3 = -90`
`(2,8 . x - 32) = -90 xx (-2/3)`
`2,8 . x = 60 + 32`
`2,8 . x = 92`
`x = 92 : 2,8`
`x = 92/2,8`
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x, biết:
(2,8.x-32):2/3=90
=>(2,8.x-32) =90.2/3
=>2,8.x-32 =60
=>2,8.x =60+32
=>x =92:2,8
=>x =
Í mà đề sai rồi, xem lại giúp mk nha !
K viết lại đề là:
2,8 x X - 32 = 90 x \(\frac{2}{3}\)
2,8 x X - 32 = \(\frac{180}{3}=60\)
2,8 x X = 60 + 32
2,8 x X = 92
x = 92 : 2,8
x = gần 32 bn nhé, có thể viết thành phân số nếu bn muốn\(\frac{920}{28}=\frac{230}{7}\)
Hội con 🐄 chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
( 2,8x - 32 ) : 2/3 = -90
2,8x - 32 = -90.2/3
2,8x - 32 = -60
2.8x = -60+32
2.8x = -28
x = -10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5+\frac{4}{7}\right):x=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\left[5+\frac{4}{7}\right]:x=13\)
\(\Rightarrow\left[\frac{5\cdot7+4}{7}\right]:x=13\)
\(\Rightarrow\frac{39}{7}:x=13\)
\(\Rightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\cdot\frac{1}{1}=\frac{3}{7}\)
\(\left[2,8\cdot x+32\right]:\frac{2}{3}=90\)
\(\Rightarrow\left[2,8\cdot x+32\right]=90\cdot\frac{2}{3}\)
\(\Rightarrow\left[2,8\cdot x+32\right]=60\)
\(\Rightarrow2,8\cdot x+32=60\)
\(\Rightarrow2,8\cdot x=60-32\)
\(\Rightarrow2,8\cdot x=28\)
Đến đây dễ rồi :v
![](https://rs.olm.vn/images/avt/0.png?1311)
(2,8.x-32):2/3=-90
=>2,8.x-32=90x2/3
=>2,8.x-32=60
=>2,8.x=60+32
=>2,8.x=92
=>x=92:2,8
=>x=230/7
( 2,8.x - 32 ) : \(\frac{2}{3}\) = - 90
= 2,8.x - 32 = 90. \(\frac{2}{3}\)
= 2,8.x - 32 = 60
= 2,8.x = 60 + 32
= 2,8.x = 92
x = 92 : 2,8
x = \(\frac{230}{7}\)
Vậy x bằng \(\frac{230}{7}\)