Tìm số nguyên x biết 2x+|x|=3x
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\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{1;0;4;-3\right\}\)
\(\Leftrightarrow\left(6x-10\right)⋮\left(2x-1\right)\\ \Leftrightarrow\left[3\left(2x-1\right)-7\right]⋮\left(2x-1\right)\\ \Leftrightarrow2x-1\inƯ\left(-7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow2x\in\left\{-6;0;2;8\right\}\\ \Leftrightarrow x\in\left\{-3;0;1;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-[-3x+2(45-x-9x+21-2x)+4x=55
-[-3x+2(24-12x)+4x]=55
-(-3x+48-24x+4x)=55
-(48-23x)=55
23x=55+48
x=103/23
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`(4 + 2x) * (9 - 3x) = 0`
TH1: `4 + 2x = 0 => 2x = -4 => x = -2`
TH2: `9 - 3x = 0 => 3x = 9 => x = 3`
Vậy, `x \in {-2; 3}`
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`(3x - 19)^3 = 125`
`=> (3x - 19)^3 = 5^3`
`=> 3x - 19 = 5`
`=> 3x = 24`
`=> x = 8`
Vậy, `x = 8.`
![](https://rs.olm.vn/images/avt/0.png?1311)
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a, \(\left(x-1\right).\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b, \(\left(2x-4\right).\left(3x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\3x+9=0\end{matrix}\right.\left[{}\begin{matrix}2x=4\\3x=-9\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a) TH1: x-1=0 => x=1
TH2: x+2=0 => x=-2
b) TH1: 2x-4=0 <=> 2x= 4 <=> x=2
TH2: 3x+9=0 <=> 3x=-9 <=> x= -3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)