Question

Proszę rozwiązać to zadanie.

Answer 1

[tex]x^2+2\geq\dfrac{6}{3-x}\Big|\cdot (3-x)^2\qquad(x\not=3)\\(x^2+2)(3-x)^2\geq6(3-x)\\(x^2+2)(9-6x+x^2)\geq18-6x\\9x^2-6x^3+x^4+18-12x+2x^2-18+6x\geq0\\x^4-6x^3+11x^2-6x\geq0\\x(x^3-6x^2+11x-6)\geq0\\x_0=0\\\\x^3-6x^2+11x-6=0\\x^3-2x^2-4x^2+8x+3x-6=0\\x^2(x-2)-4x(x-2)+3(x-2)=0\\(x^2-4x+3)(x-2)=0\\\\x-2=0\\x_0=2\\\\x^2-4x+3=0\\x^2-x-3x+3=0\\x(x-1)-3(x-1)=0\\(x-3)(x-1)=0\\x_0=3 \vee x_0=1[/tex]

[tex]x\in(-\infty,0\rangle\cup\langle1,2\rangle\cup(3,\infty)[/tex]