$(x+1)^2>[6-(2-3x)]1/3x-3$

A cura di: Francesco Speciale

Svolgimento:
$(x+1)^2>[6-(2-3x)]1/3x-3$
$x^2+1+2x>[6-2+3x]1/3x-3$
$x^2+1+2x>(4+3x)1/3x-3$
$x^2+1+2x-4/3x-x^2+3>0$
Semplificando si ha
$(2-4/3)x>-4$
$2/3x>-4=>x>(-4)*3/2=>x>-6$.