1)
[tex](x-3)(2x+5)\leq(2x-6)(2x-1)\\2x^2+5x-6x-15\leq4x^2-2x-12x+6\\2x^2-4x^2-x+14x-15-6\leq0\\-2x^2+13x-21\leq0\\\Delta=(13)^2-4*(-2)*(-21)\\\Delta=169-168\\\Delta=1\\\sqrt{\Delta}=1\\x_1=\frac{-13-1}{-4}=\frac{-14}{-4}=\frac{14}4=3\frac24=3\frac12=\frac72\\x_2=\frac{-13+1}{-4}=\frac{-12}{-4}=3\\a=-2 / a<0 - \text{ramiona paraboli skierowane w dol}[/tex]
x∈(-∞; 3>∪<3.5; ∞)
2)
[tex]2(1-3x)(3x-9)\geq x^2-9\\(2-6x)(3x-9)\geq x^2-9\\6x-18-18x^2+54x-x^2+9\geq0\\-19x^2+60x-9\geq0\\\Delta=60^2-4*(-19)*(-9)\\\Delta=3600-684=2916\\\sqrt{\Delta}=54\\x_1=\frac{-60-54}{-38}=\frac{-114}{-38}=3\\x_2=\frac{-60+54}{-38}=\frac{-6}{-38}=\frac{3}{19}\\a=-19 / a<0 - \text{ramiona paraboli skierowane w dol}\\x\in<\frac3{19}; 3>[/tex]
3)
[tex]\frac{x-\frac{x^2-4}2}5\geq\frac{3-\frac{(x+2)^2}4}2+\frac18\\\frac15*(x-\frac{x^2-4}2)\geq \frac12(3-\frac{x^2+4x+4}4)+\frac18\\\frac15x-\frac{x^2-4}{10}\geq \frac32-\frac{x^2+4x+4}8+\frac18\\\frac2{10}x-\frac{x^2-4}{10} \geq \frac{12-(x^2+4x+4)+1}{8}\\\frac{2x-(x^2-4)}{10}\geq\frac{12-x^2-4x-4+1}8\\\frac{-x^2+2x+4}{10} \geq \frac{-x^2-4x+9}8 /*80\\8(-x^2+2x+4)\geq 10(-x^2-4x+9)\\-8x^2+16x+32\geq -10x^2-40x+90\\-8x^2+10x^2+16x+40x+32-90\geq0\\2x^2+56x-58 \geq 0\\[/tex]
[tex]\Delta=(56)^2-4*2*(-58)\\\Delta=3136+464\\\Delta=3600\\\sqrt{\Delta}=60\\x_1=\frac{-56-60}{4}=\frac{-116}4=-29\\x_2=\frac{-56+60}4=\frac44=1\\a=2 / a>0 - \text{ramiona paraboli skierowane w gore}[/tex]
x∈(-∞; -29>∪<1; ∞)
