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Ta có
$4 + \sqrt{15}=\frac{1}{2}(8 + 2\sqrt{15})=\frac{1}{2}(\sqrt 5 + \sqrt 3)^2$
$ \sqrt{10} - \sqrt {6}=\sqrt{2}(\sqrt 5 - \sqrt 3)$
$\sqrt{4 - \sqrt{15}}=\sqrt{\frac{1}{2}(8 - 2\sqrt{15})}=\sqrt{\frac{1}{2}(\sqrt 5 - \sqrt 3)^2}=\frac{1}{\sqrt 2}(\sqrt 5 - \sqrt 3)$
Vậy
$(4+\sqrt{15}) ( \sqrt{10} - \sqrt {6}) \sqrt {4- \sqrt15} =\frac{1}{2}(\sqrt 5 + \sqrt 3)^2(\sqrt 5 - \sqrt 3)^2=\frac{1}{2}(5-3)^2=2$. đpcm.
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