\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{1;0;4;-3\right\}\)

\(\Leftrightarrow\left(6x-10\right)⋮\left(2x-1\right)\\ \Leftrightarrow\left[3\left(2x-1\right)-7\right]⋮\left(2x-1\right)\\ \Leftrightarrow2x-1\inƯ\left(-7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow2x\in\left\{-6;0;2;8\right\}\\ \Leftrightarrow x\in\left\{-3;0;1;4\right\}\)

x=103/23

-[-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

Xem lại dùm mink nha

`\color{grey}\text{#071931}`

`(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}`

____

`(3x - 19)^3 = 125`

`=> (3x - 19)^3 = 5^3`

`=> 3x - 19 = 5`

`=> 3x = 24`

`=> x = 8`

Vậy, `x = 8.`

a) Không tồn tại x

b) x = -7

c) x = 2

d) x = 5

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

a) ( x - 1) . ( x + 2 ) = 0 

=> x - 1 = 0 => x = 1

     x + 2 = 0     x = -2 

vậy x ϵ { 1; -2 }

b) ( 2x - 4 ) . ( 3x + 9 ) = 0

=> 2x - 4  = 0   => 2x = 4  => x = 2

     3x + 9 = 0        3x = -9      x = -3

vậy x ϵ { 2 ; -3 {

P/S : phần mình suy ra thì bn đóng ngoặc vuông to rồi mới ghi phép tính nhé!