pertanyaan tentang nilai mutlak no.12

2|x + 1| < 1|x -1|
|2x + 2| < |x - 1|
(2x + 2 + x - 1)(2x + 2 - x + 1) < 0
(3x + 1)(x + 3) < 0

3x = -1
x = -1/3

x = -3

HP = {x | -3 < x < -1/3} 
(A)

Maaf kalau salah

2|x+1|<|x-1|
|2x+2|²<|x-1|²
|2x+2|²-|x-1|²<0
(2x+2+x-1)(2x+2-x+1)<0
(3x+1)(x+3)<0
3x²+9x+x+3<0
3x²+10x+3<0(negatif)
faktorkan
(x+3)(3x+1)<0
batas: x=-3  ⇒x>-3(negatif)
         x=-1/3⇒x<-1/3(negatif)
hp={x|-3<x<-1/3,eR}⇒⇒A.