Cho lăng trụ  $ABC.A'B'C' $ có $ A'.ABC$ là chóp tam giác đều. $AB=a$. Gọi $\alpha$ là góc giữa  $(A'BC)$ và $(C'B'BC)$. Tính  theo $a$ thể tích chóp  $A'BCC'B'$. Biết  $\cos \alpha =1/\sqrt 3$
dạng này khó thế –  duachua.no1 07-11-12 10:48 PM
mình chưa làm dạng này bh cả –  yeuhoahocneu 07-11-12 08:53 PM
thía á, thầy giáo t cũng cho hôm qua –  luffykunneu 07-11-12 08:09 PM
bài này cô mình cũng cho –  banhquykeomut 07-11-12 08:07 PM

bị xóa vì bị trùng bởi Đức Vỹ 24-10-13 03:56 PM

Câu hỏi này có nội dung hoàn toàn trùng với câu hỏi đã có.

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