(x-3)÷(x^2-8x+12)больше или равно 0 (x^2-3x-4)÷(x^2+x-6)>0

(x-3)÷(x^2-8x+12)≥0
ОДЗ
(x^2-8x+12)=0
D=64-4*12=16
x₁=(8-4)/2=2
x₂=(8+4)/2=6

$\frac{x-3}{(x-2)(x-6)} \geq 0$
x∈(2; 3]∨(6; +∞)

(x^2-3x-4)÷(x^2+x-6)>0
ОДЗ
х²+х-6=0
D=1+6*4=25
x₁=(-1-5)/2=-3
x₂=(-1+5)/2=2

$\frac{x^2-3x-4}{x^2+x-6} \ \textgreater \ 0 \\ \frac{x^2-4x+x-4}{x^2+x-6} \ \textgreater \ 0 \\ \frac{x^2-4x+x-4}{x^2+x-6} \ \textgreater \ 0 \\ \frac{(x-4)(x+1)}{(x+3)(x-2)} \ \textgreater \ 0 \\$
x∈(-∞; -3)∨(-1;2)∨(4; +∞)