Verified answer

Объяснение:

[tex]\left \{ {{x^2-3x+9 > 0} \atop {x^2\leq 36}} \right.\ \ \ \ \left \{ {{} \atop {x^2-6^2\leq 0}} \right. \ \ \ \ \ \left \{ {{x^2-2*x*1,5+1.5^2-1,5^2+9 > 0} \atop {(x+6)*(x-6)\leq 0}} \right.\ \ \ \ \ \left \{ {{(x-1,5)+6,75 > 0} \atop {(x+6)*(x-6)\leq 0}} \right. \\(x-1,5)^2+6,75 > 0\ \ \ \ \ \Rightarrow\\x\in(-\infty;+\infty).\\(x+6)*(x-6)\leq 0\\[/tex]

-∞__+__-6__-__6__+__+∞

x∈[-6;6].           ⇒

Ответ: x∈[-6;6].