Question

розвʼяжіть нерівність:

Answer 1

$\frac{x^3+x^2+3x+3}{x^2-6x+7}\leq 0\Leftrightarrow \frac{x^2(x+1)+3(x+1)}{\left ( x-3+\sqrt{2} \right )\left ( x-3-\sqrt{2} \right )}\leq 0\Leftrightarrow \frac{(x+1)\left ( x^2+3 \right )}{\left ( x-3+\sqrt{2} \right )\left ( x-3-\sqrt{2} \right )}\leq 0 x\in (-\infty ,-1]\cup \left ( 3-\sqrt{2},3+\sqrt{2} \right )$