a. $\dfrac{6 - 3\sqrt{6}}{6} = \dfrac{\sqrt{6}(\sqrt{6} - 3}{(\sqrt{6})^2} = \dfrac{\sqrt{3}(\sqrt{2} - \sqrt{3})}{\sqrt{2}.\sqrt{3}} = \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2}}$
b. $\dfrac{\sqrt{42} - 6}{\sqrt{21} - \sqrt{18}} = \dfrac{\sqrt{6}(\sqrt{7} - \sqrt{6})}{\sqrt{3}(\sqrt{7} - \sqrt{7})} = \dfrac{\sqrt{2}.\sqrt{3}}{\sqrt{3}} = \sqrt{2}$
c. $\sqrt{\sqrt{5} - \sqrt{3 - \sqrt{29 - 6\sqrt{20}}}}$
$= \sqrt{\sqrt{5} - \sqrt{3 - \sqrt{(2\sqrt{5} - 3)^2}}}$
$= \sqrt{\sqrt{5} - \sqrt{3 - (2\sqrt{5} - 3)}}$
$= \sqrt{\sqrt{5} - \sqrt{6 - 2\sqrt{5}}}$
$= \sqrt{\sqrt{5} - \sqrt{(\sqrt{5} - 1)^2}}$ $= \sqrt{\sqrt{5} - \sqrt{5} + 1} = \sqrt{1} = 1$
d. $(4 + \sqrt{15})(\sqrt{10} - \sqrt{6})\sqrt{4 - \sqrt{15}}$
$= (4 + \sqrt{15}).\sqrt{2}(\sqrt{5} - \sqrt{3})\sqrt{4 - \sqrt{15}}$
$= (4 + \sqrt{15})(\sqrt{5} - \sqrt{3})\sqrt{2(4 - \sqrt{15})}$
$= (4 + \sqrt{15})(\sqrt{5} - \sqrt{3})\sqrt{(\sqrt{5} - \sqrt{3})^2}$
$= (4 + \sqrt{15})(\sqrt{5} - \sqrt{3})(\sqrt{5} - \sqrt{3})$
$= (4 + \sqrt{15})(5 + 3 - 2\sqrt{15}) = (4 + \sqrt{15}).2.(4 - \sqrt{15}) = 2.(16 - 15) = 2$
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(Câu d có sửa đề chút nhé!)
