a)
[tex]2\sqrt3=\sqrt{12}=3,...\Rightarrow 2\sqrt3-5<0\Rightarrow |2\sqrt3-5|=-(2\sqrt3-5)=5-2\sqrt3 \\\\\displaystyle x=\frac{4}{4-\sqrt{12}}+|2\sqrt3-5|=\frac{4(4+\sqrt{12})}{4^2-(\sqrt{12})^2}+5-2\sqrt3=\frac{4(4+\sqrt{12})}{4}+5-2\sqrt3=4+2\sqrt3+5-2\sqrt3=9\in\mathbb{Z}[/tex]
b)
[tex]\displaystyle y=\frac{1}{\sqrt2}+\frac{15}{\sqrt{18}}=\frac{\sqrt2}{2}+\frac{15\sqrt{18}}{18}=\frac{9\sqrt2+15\cdot3\sqrt2}{18}=\frac{\sqrt2+5\sqrt2}{2}=3\sqrt2\\\\9>2\cdot3\sqrt2\\9>6\sqrt2\\9>\sqrt{72}\\\sqrt{81}>\sqrt{72}\\81>72\rightarrow x>2y[/tex]
