[tex]a)\\x\sqrt3+3\geq x+1\\x\sqrt3-x \geq 1-3\\x(\sqrt3-1)\geq -2 /:(\sqrt3-1)\\x \geq \frac{-2}{\sqrt3-1}*\frac{\sqrt3+1}{\sqrt3+1}\\x \geq \frac{-2(\sqrt3+1)}{3-1}\\x \geq \frac{-2\sqrt3-2}{2}\\x \geq \frac{2(-\sqrt3-1)}2\\x \geq -\sqrt3-1[/tex]
[tex]b)\\-(x+2)(6+x) \geq 0\\-(6x+x^2+12+2x) \geq 0\\-(x^2+8x+12)\geq 0\\-x^2-8x-12 \geq 0\\\Delta=(-8)^2-4*(-1)*(-12)\\\Delta=64-48\\\Delta=16\\\sqrt{\Delta}=4\\x_1=\frac{8-4}{-2}=\frac{4}{-2}=-2\\x_2=\frac{8+4}{-2}=\frac{12}{-2}=-6\\f(x) \geq 0 \to x\in < -6; -2 >[/tex]
[tex]c)\\9x^2-12 \leq 0\\\Delta=0^2-4*9*(-12)\\\Delta=432\\\sqrt{\Delta}=\sqrt{432}=12\sqrt3\\\\x_1=\frac{-12\sqrt3}{18}=-\frac{2\sqrt3}3\\x_2=\frac{12\sqrt3}{18}=\frac{2\sqrt3}3\\x\in < -\frac{2\sqrt3}3; \frac{2\sqrt3}3 >[/tex]
[tex]d)\\x^3-2x^2-9x=0\\x(x^2-2x-9)=0\\x=0\\\Delta=(-2)^2-4*1*(-9)\\\Delta=4+36\\\Delta=40\\\sqrt{\Delta}=\sqrt{40}=2\sqrt{10}\\\\x_1=\frac{2-2\sqrt{10}}{2}=\frac{2(1-\sqrt{10})}2=1-\sqrt{10}\\x_2=\frac{2+2\sqrt{10}}2=1+\sqrt{10}\\x_3=0[/tex]
[tex]e) \\\frac{(x^2-4)(x+7)}{(x+2)(x-3)}=0\\\frac{(x-2)(x+2)(x+7)}{(x+2)(x-3)}=0\\\frac{(x-2)(x+7)}{x-3}=0\\x-3 \neq 0\\x \neq 3\\\\(x-2)(x+7)=0\\x-2=0\\x=2\\\\x+7=0\\x=-7[/tex]
[tex]f)\\\\6x=3x^2\\-3x^2+6x=0\\\Delta=6^2-4*(-3)*0\\\Delta=6^2=36\\\sqrt{\Delta}=\sqrt{6^2}=6\\x_1=\frac{-6-6}{-6}=\frac{-12}{-6}=2\\x_2=\frac{-6+6}{-6}=0[/tex]
