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= (2x+3)² > (x+1)²
= (2x+3)² - (x+1)² > 0
= {(2x+3)-(x+1)}{(2x+3)+(x+1)}
= (x+2) > 0 (3x+4) > 0
x > -2 x <
Hp : {x|x>-2 atau x< x ∈ R}
Semoga bermanfaat :))
= (3x+4) (x+2) > 0
batas
3x+4 = 0 atau x+2 = 0
3x = -4 x= -2
x = -4/3
HP= {x|x< -2, atau x> -4/3,x∈R}