求 $\sqrt{\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}}$ 的值，并将你的答案写成 $a + b\sqrt{6}$ 的形式。 不使用计算器，计算 $(2 - \sqrt[3]{4})\left[4 + (2\times\sqrt[3]{4}) + \sqrt[3]{16}\right]$ 的值。

1. Rationalise and simplify the expression inside the square root.

$\frac{5+2\sqrt6}{5-2\sqrt6}\times \frac{5+2\sqrt6}{5+2\sqrt6}=\frac{(5+2\sqrt6)^2}{5^2-(2\sqrt6)^2}$
$=\frac{(5+2\sqrt6)^2}{25-24}=(5+2\sqrt6)^2$

2. Take the square root of the perfect square surd expression.

$\sqrt{\frac{5+2\sqrt6}{5-2\sqrt6}}=\sqrt{(5+2\sqrt6)^2}$
Since $5+2\sqrt6>0$,
$\sqrt{(5+2\sqrt6)^2}=5+2\sqrt6.$

1. Recognise the bracket as matching the difference of cubes identity.

Let $x=\sqrt[3]{4}$. Then $x^2=\sqrt[3]{16}$, so
$(2-\sqrt[3]{4})\left[4+(2\times\sqrt[3]{4})+\sqrt[3]{16}\right]=(2-x)(4+2x+x^2).$

2. Apply the difference of cubes identity and evaluate.

Using $(a-b)(a^2+ab+b^2)=a^3-b^3$,
$(2-x)(4+2x+x^2)=2^3-x^3=8-(\sqrt[3]{4})^3=8-4=4.$