Verified answer

Объяснение:

[tex]\left \{ {{x^2-3x+9 > 0} \atop {x^2\leq 36}} \right. \ \ \ \ \left \{ {{x^2-2*x*1,5+1,5^2-1.5^2+9 > 0} \atop {x^2-6^2\leq 0}} \right. \ \ \ \ \ \left \{ {{(x-1,5)^2-2,25+9 > 0} \atop {(x+6)*(x-6)\leq 0}} \right. \\[/tex]

[tex]\left \{ {{(x-1,5)^2+6.75 > 0} \atop {x\in[-6;6]}} \right. \ \ \ \ \left \{ {{x\in(-\infty;+\infty)} \atop {x\in[-6;6]}} \right. .[/tex]

-∞__+__-6__-__6__+__+∞          ⇒

Ответ: x∈[-6;6].