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	Các bình luận về: Cho tứ diện ABCD có \(AB = a; AC = BC = AD = BD =
\frac{{a\sqrt 3 }}{2}\). Gọi M, Nlà trung điểm của AB, CD. Góc giữa hai mặt phẳng (ABD) ; (ABC) là \(\alpha\) . Tính \(cos \alpha\) biết mặt cầu đường kính MN tiếp xúc với cạnh AD	</title>
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