(x-3)÷(x^2-8x+12)больше или равно 0 (x^2-3x-4)÷(x^2+x-6) больше 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 [latex] \frac{x-3}{(x-2)(x-6)} \geq 0 [/latex] 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 [latex] \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 \\ [/latex] x∈(-∞; -3)∨(-1;2)∨(4; +∞)