Chứng minh rằng với $ \frac{3}{2} \ge x \ge 0$ thì với mọi tam giác ABC có 3 cạnh $a;b;c$ sao cho $xa;c.cosC;xb $ theo thứ tự lập thành cấp số cộng thì ta luôn có:
$1 \ge cosC \ge \sqrt{\frac{9}{16}-\frac{x^2}{4}}-\frac{1}{4}$    hoặc:  $-1 \le cosC \le -\sqrt{\frac{9}{16}-\frac{x^2}{4}}-\frac{1}{4}$
đề quá hay –  luffykunneu 14-11-12 08:25 PM
nhìn đã loạn mắt rồi –  banhquykeomut 14-11-12 08:24 PM

bị xóa vì bị trùng bởi Đức Vỹ 24-10-13 03:49 PM

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

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