So sánh cặp số sau: 

- Hướng dẫn giải

Ta có

$$\begin{gathered} A=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2} \\ =\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)} \\ =\frac{15\left(\sqrt{6}-1\right)}{6-1}+\frac{4\left(\sqrt{6}+2\right)}{6-4} \\ =3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right) \\ =5\sqrt{6}+1 \end{gathered}$$

$$\begin{gathered} B=\frac{12}{3-\sqrt{6}}+\sqrt{6}=\frac{12.\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)} \\ =\frac{12.\left(3+\sqrt{6}\right)}{9-6}=4\left(3+\sqrt{6}\right) \end{gathered}$$

Vậy ta được A < B.