Odpowiedź :
1.
[tex]2x^{2}-9 \leq 0 \ \ |:2\\\\x^{2} - \frac{9}{2}\leq 0\\\\(x+\sqrt{\frac{9}{2}})(x-\sqrt{\frac{9}{2}})\leq 0\\\\(x+\frac{\sqrt{9}}{\sqrt{2}})(x-\frac{\sqrt{9}}{\sqrt{2}}) \leq 0\\\\(x+\frac{3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}})(x-\frac{3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}})\leq 0\\\\(x+\frac{3\sqrt{2}}{2})(x-\frac{3\sqrt{2}}{2})\leq 0\\\\\\M. \ zerowe:\\\\x+\frac{3\sqrt{2}}{2} = 0 \ \ \vee \ \ x -\frac{3\sqrt{2}}{2} = 0\\\\x = -\frac{3\sqrt{2}}{2} \ \ \vee \ \ x = \frac{3\sqrt{2}}{2}[/tex]
a > 0, to parabola zwrócona jest ramionami do góry, wówczas:
[tex]\boxed{x \in \langle -\frac{3\sqrt{2}}{2}; \frac{3\sqrt{2}}{2}\rangle}[/tex]
2.
[tex]log_{3}2 + log_{3}13,5 = log_{3}(2\cdot13,5) = log_{3}27 = log_{3}3^{3} = \boxed{3}[/tex]