回答所有问题。 (a) 将 $\dfrac{6\sqrt{2}-9}{\sqrt{2}-1}$ 表示为 $a+b\sqrt{2}$ 的形式，其中 $a$ 和 $b$ 为整数。 (b) $\dfrac{6\sqrt{2}-9}{\sqrt{2}-1}$ 是方程 $x^{2}+px+q=0$ 的一个根，其中 $p$ 和 $q$ 为整数。 利用 (a) 部分的结果，求 $p$ 和 $q$ 的值。

1. Evaluate the expression from part (a) to find the root

$\frac{6\sqrt{2}-9}{\sqrt{2}-1} = \frac{(6\sqrt{2}-9)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{12 + 6\sqrt{2} - 9\sqrt{2} - 9}{2 - 1} = 3 - 3\sqrt{2}$

2. Substitute the root into the given quadratic equation

$(3 - 3\sqrt{2})^2 + p(3 - 3\sqrt{2}) + q = 0$

3. Expand the equation and group rational and irrational terms

$9 - 18\sqrt{2} + 18 + 3p - 3p\sqrt{2} + q = 0$
$(27 + 3p + q) - (18 + 3p)\sqrt{2} = 0$

4. Compare coefficients to find $p$ and $q$

Since $p$ and $q$ are integers, the rational and irrational parts must independently equal zero.
$18 + 3p = 0 \implies p = -6$
$27 + 3(-6) + q = 0 \implies 27 - 18 + q = 0 \implies q = -9$