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Verified answer
Bab Nilai Mutlak3|x+3| < |x-3| [kuadratkan ke-2 ruas untuk menghilangkan tanda mutlak]
( 3(x+3) )² < (x-3)²
(3x + 9)² < (x - 3)²
(3x + 9)² - (x - 3)² < 0 [Gunakan (a² - b²)=(a + b)(a - b) ]
(3x + 9 + x - 3)(3x + 9 -x + 3) < 0
(4x + 6)(2x + 12)<0
Pembuat 0
4x + 6=0 ---- 2x + 12=0
x=-6/4 ---- x=-6
x=-3/2
Karena < 0,berarti HP-nya interval
HP:{x|-6 < x < -3/2, x ∈ R}