Ta có :

a.   $2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}$ = $27-5\sqrt{3x}$

Vậy $2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x} = 27-5\sqrt{3x}$

b.  $3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28$

= $3\sqrt{2x}-5\sqrt{2^{2}.2x}+7\sqrt{3^{2}.2x}+28$

= $3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28$

= $14\sqrt{2x}+28$

Vậy $3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28=14\sqrt{2x}+28$