Question

1. |5/x-3| < 2 2. |5/x-3| > 2

Answer 1

1. Ι5/х-3Ι<2  ОДЗ: х≠0
При раскрытии модуля имеем систему неравенств:
5/х-3<2          5-3х<2x          x>1   
-5/x+3<2       -5+3x<2x         x<5  ⇒  x∈(1;5).
2. Ι5/3x-3Ι>2 
x<1  
x>5   ⇒  x∉, то есть неравенство не имеет решений.

Answer 2

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