[tex]\it a.\ \ \sqrt5-\dfrac{5}{\sqrt5}=\sqrt5-\dfrac{\sqrt5\cdot\sqrt5}{\sqrt5}=\sqrt5-\sqrt5=0\in\mathbb{Q}\\ \\ \\ b.\ \ \dfrac{6}{\sqrt6}-\sqrt6=\dfrac{\sqrt6\cdot\sqrt6}{\sqrt6}-\sqrt6=\sqrt6-\sqrt6=0\in\mathbb{Q}\\ \\ \\ c.\ \ \dfrac{6}{\sqrt{12}}+\dfrac{\sqrt3-3}{\sqrt3}=\dfrac{\ \ 6^{(2}}{2\sqrt3}+\dfrac{\sqrt3-3}{\sqrt3}=\dfrac{3+\sqrt3-3}{\sqrt3}=\dfrac{\sqrt3}{\sqrt3}=1\in\mathbb{Q}[/tex]
[tex]\it d.\ \ \dfrac{5}{\sqrt{125}}-\dfrac{\sqrt{36}}{\sqrt{180}}=\dfrac{\sqrt{25}}{\sqrt{125}}-\dfrac{\sqrt{36}}{\sqrt{180}}=\dfrac{1}{\sqrt5}-\dfrac{1}{\sqrt5}=0\in\mathbb{Q}[/tex]
