4x\\\\x^2+5x-4x-2>0\; \; \to \\\\x^2+x-2>0\; ,\; \; x_1=-2\; ,\; x_2=1\; (teorema\; Vieta)\\\\(x+2)(x-1)>0\quad +++(-2)---(1)+++\\\\x\in (-\infty ,-2)\cup (1,+\infty )" alt=" 1)\; \; 11-(x+1)^2\geq x\\\\11-x^2-2x-1\geq x\\\\x^2+3x-10\leq 0\; ,\; \; x_1=-5\; ,\; x_2=2\; (teorema\; Vieta)\\\\znaki:\quad +++[-5\, ]---[\, 2\, ]+++\\\\x\in [-5,2\, ]\\\\2)\; \; (2x-8)^2-4x(2x-8)\geq 0\\\\4x^2-32x+64-8x^2+32x\geq 0\\\\-4x^2+64\geq 0\; |:(-4)\; \; \to \; \; x^2-16\leq 0\\\\(x-4)(x+4)\leq 0\; ,\qquad +++[-4\, ]---[\, 4\, ]+++\\\\x\in [-4,4\, ]\\\\3)\; \; x(x+5)-2>4x\\\\x^2+5x-4x-2>0\; \; \to \\\\x^2+x-2>0\; ,\; \; x_1=-2\; ,\; x_2=1\; (teorema\; Vieta)\\\\(x+2)(x-1)>0\quad +++(-2)---(1)+++\\\\x\in (-\infty ,-2)\cup (1,+\infty )" align="absmiddle" class="latex-formula">
\frac{x^2}{2}-4x+5\frac{1}{2}\; |\cdot 2\\\\2x>x^2-8x+11\; ,\\\\x^2-10x+11<0\; ,\; D/4=5^2-11=14\; ,\; \; x_{1,2}=5\pm \sqrt{14}\\\\znaki:\quad +++(5-\sqrt{14})---(5+\sqrt{14})+++\\\\x\in (5-\sqrt{14}\; ;\; 5+\sqrt{14}) " alt=" 4)\; \; \frac{1}{3}x^2+3x+6<0\; |\cdot 3\\\\x^2+9x+18<0\; ,\; \; x_1=-6\; ,\; x_2=-3\; \; (teorema\; Vieta)\\\\(x+6)(x+3)<0\quad ++(-6)---(-3)+++\\\\x\in (-6,-3)\\\\5)\; \; x>\frac{x^2}{2}-4x+5\frac{1}{2}\; |\cdot 2\\\\2x>x^2-8x+11\; ,\\\\x^2-10x+11<0\; ,\; D/4=5^2-11=14\; ,\; \; x_{1,2}=5\pm \sqrt{14}\\\\znaki:\quad +++(5-\sqrt{14})---(5+\sqrt{14})+++\\\\x\in (5-\sqrt{14}\; ;\; 5+\sqrt{14}) " align="absmiddle" class="latex-formula">