Question

Answer 1

a: =>x+3=-20

hay x=-23

b: =>4x+12=0

hay x=-3

c: =>\(x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

Answer 2

a. x+3 =-20

=> x= -23

b. 4x+12=0 hoặc \(x^2+4=0\) 

=> x=-3,\(x^2=-4\) ( vô lí)

=> x=-3

Answer 3

a, -120:(x+3) = 6

<=> x + 3 =-20 <=> x =-23

b, (4x+12)(x^2+4)=0

<=> x = -12/4 = -3 

do x^2 + 4 > 0 Vậy x = -3 

c, \(2x-5⋮x-1\Leftrightarrow2\left(x-1\right)-3⋮x-1\Rightarrow x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

x-1	1	-1	3	-3
x	2	0	4	-2