Đáp án:
$d)(3\sqrt{2}+\sqrt{6})\sqrt{6-\sqrt{27}}$
$=3\sqrt{2}.\sqrt{6-\sqrt{27}}+\sqrt{6}.\sqrt{6-\sqrt{27}}$
$=3\sqrt{12-2\sqrt{27}}+\sqrt{36-6\sqrt{27}}$
$=3\sqrt{(\sqrt{9})^2-2\sqrt{27}+(\sqrt{3})^2}+\sqrt{9-6\sqrt{27}+(\sqrt{27})^2}$
$=3\sqrt{(\sqrt{9}-\sqrt{3})^2}+\sqrt{(3-\sqrt{27})^2}$
$=3|3-\sqrt{3}|+|3-3\sqrt{3}|$
$=3(3-\sqrt{3})+(3\sqrt{3}-3)$
$=9-3\sqrt{3}+3\sqrt{3}-3$
$=9-3=6$
