Đáp án:
$2\sqrt[]{6} = \sqrt[]{2^2 .6} = \sqrt[]{4.6} = \sqrt[]{24}$
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$-\dfrac{3}{4}ab\sqrt[]{32a} = -\sqrt[]{(\dfrac{3}{4}ab)^2 . 32a}$
$=-\sqrt[]{\dfrac{9}{16}a^2b^2 . 32a} = -\sqrt[]{\dfrac{288}{16}a^3b^2} (a ≥ 0 ) $