Odpowiedź :
[tex]-x^{2}+10x-1>8\\\\-x^{2}+10x-1-8 > 0\\\\-x^{2}+10x-9 > 0 \ \ /\cdot(-1)\\\\\underline{x^{2}-10x+9 < 0}[/tex]
[tex]a = 1, \ b = -10, \ c = 9\\\\M. \ zerowe\\\\\Delta= b^{2}-4ac = (-10)^{2}-4\cdot9 = 100-36 = 64\\\\\sqrt{\Delta} = \sqrt{64} = 8\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-10)-8}{2\cdot1} = \frac{2}{2} = 1\\\\x_2 =\frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-10)+8}{2} = \frac{18}{2} = 9\\\\a > 0, \ ramiona \ paraboli \ skierowane \ do \ gory\\\\x \in (1; 9)[/tex]
[tex]-x^2+10x-1>8\\x^2-10x+9<0\\x^2-10x+25-16<0\\(x-5)^2<16\\x-5<4 \wedge x-5>-4\\x<9 \wedge x>1\\x\in(1,9)[/tex]