(х+2)(х-3)≤х² (4х+1)/7 - х/2 >0
х²-3х+2х-6-х²≤0 [2(4х+1)-7х]/14>0
-х-6≤0 -х≤6 (* -1) х≥-6 (8х+2-7х)/14 >0 (*14)
х+2>0 х>-2
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х∈(-2;+∞)
Ответ: х∈(-2;+∞)