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Đáp án:

$A=\dfrac{1}{2}.$

Giải thích các bước giải:

$A=\sqrt{\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\dfrac{2}{\sqrt{3}-1}\\ =\sqrt{\dfrac{\sqrt{10}(2+\sqrt{3})-\sqrt{2}(2+\sqrt{3})}{2\sqrt{10}-2\sqrt{2}}}.\dfrac{\sqrt{3}-1}{2}\\ =\sqrt{\dfrac{(\sqrt{10}-\sqrt{2})(2+\sqrt{3})}{2(\sqrt{10}-\sqrt{2})}}.\dfrac{\sqrt{3}-1}{2}\\ =\sqrt{\dfrac{2+\sqrt{3}}{2}}.\dfrac{\sqrt{3}-1}{2}\\ =\sqrt{\dfrac{4+2\sqrt{3}}{4}}.\dfrac{\sqrt{3}-1}{2}\\ =\sqrt{\dfrac{3+2\sqrt{3}+1}{4}}.\dfrac{\sqrt{3}-1}{2}\\=\sqrt{\dfrac{(\sqrt{3}-1)^2}{4}}.\dfrac{\sqrt{3}-1}{2}\\ =\dfrac{\sqrt{3}+1}{2}.\dfrac{\sqrt{3}-1}{2}\\ =\dfrac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}\\ =\dfrac{3-1}{4}\\ =\dfrac{1}{2}.$