Question

решите неравенство
1) 3x^{2}+2x-1<0
2) x^{3}-5 x^{2}+6x≥0

Answer 1

$3 x^{2} +2x-1<0 \ 3 x^{2} +2x-1=0 \ a-b+c=0 \ => x_{1} =-1, x_{2}= frac{1}{3} \ x (-1; frac{1}{3} )$

$x^{3} -5 x^{2} +6x leq 0 \ x^{3} -5 x^{2} +6x=0 \ x( x^{2} -5x+6)=0 \ x_{1} =0 \ x^{2} -5x+6=0 \ a+b+c=0 \ => x_{2} =1, x_{3}=6 \ x(- infty;0] u [1;6]$