$\textit{Đáp án + Giải thích các bước giải:}$
$\text{a) $\dfrac{3}{2}$$\sqrt{6}$ + 2$\sqrt{\dfrac{3}{2}}$ - 4$\sqrt{\dfrac{3}{2}}$}$
$\text{= $\dfrac{3}{2}$$\sqrt{6}$ + 3$\dfrac{2}{3}$$\sqrt{\dfrac{2}{3}}$ - 2 . 2$\sqrt{\dfrac{3}{2}}$}$
$\text{= $\dfrac{3}{2}$$\sqrt{6}$ + $\dfrac{2}{3}$$\sqrt{3^2.\dfrac{2}{3}}$ - 2$\sqrt{2^2.\dfrac{3}{2}}$}$
$\text{= $\dfrac{3}{2}$$\sqrt{6}$ + $\dfrac{3}{2}$$\sqrt{6}$ - 2$\sqrt{6}$}$
$\text{= $\bigg(\dfrac{3}{2}+\dfrac{2}{3}-2\bigg)$$\sqrt{6}$}$
$\text{= $\dfrac{1}{6}$$\sqrt{6}$ = $\dfrac{\sqrt{6}}{6}$}$
$\text{b) $\bigg(x + \sqrt{\dfrac{6}{x}} + \sqrt{\dfrac{2x}{3}} + \sqrt{6x}\bigg)$ : $\sqrt{6x}$}$
$\text{= $\bigg(\sqrt{x^2 . \dfrac{6}{x}} + \sqrt{\dfrac{2x . 3}{3 . 3}} + \sqrt{6}\bigg)$ : $\sqrt{6x}$}$
$\text{= $\bigg(\sqrt{6x} + \dfrac{1}{3}\sqrt{6x} + \sqrt{6x}\bigg)$ : $\sqrt{6x}$}$
$\text{= $\bigg[\sqrt{6x}\bigg(1 + 1 + \dfrac{1}{3}\bigg)\bigg]$ : $\sqrt{6x}$}$
$\text{= $\sqrt{6x}$ . $\dfrac{7}{3}$ : $\sqrt{6x}$}$
$\text{= $\dfrac{7}{3}$ = 2$\dfrac{1}{3}$}$
