Ответ №1
1) (2y+3)(y+3)=<0
y1=-3/2 y2=-3
+ - +
--------(-3)----------(-3/2)----------
x∈[-3,-3/2]
2) (x+12)(x-5)<0
x1=-12 x2=5
+ - +
-----------(-12)----------(5)--------
x∈(-12,5)
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