Odpowiedź :
Odpowiedź:
Szczegółowe wyjaśnienie:
Zadanie a
[tex]-x^{2} +3x-x+3\leq 0\\-x* (x-3) - (x-3) \leq 0\\-(x-3)*(x+1)\geq 0\\\\x-3\geq 0\\x+1\geq 0\\\\x-3\leq 0\\x+1\leq 0\\x\geq 3\\x\geq -1\\x\leq 3\\x\leq -1\\xe [ 3 + n >\\xe < - n, - 1 ]\\ZADANIE B \\x^{2} +6x+6>0\\x^{2} +6x+6=0\\x=-3+\sqrt{3}\\x=-3-\sqrt{3}\\(x-(-3+\sqrt{3}))) * ( x-(-3-\sqrt{3} )) > 0\\x=3-\sqrt{3} >0\\x+3 + \sqrt{3}>0 \\x+3-\sqrt{3} <0\\x+3+\sqrt{3} <0\\x>-3+\sqrt{3}\\x>-3\sqrt{3}\\x<-3+\sqrt{3}\\x< -3 - \sqrt{3}\\\\xe < -3 + \sqrt{3} + n >\\[/tex]
