GPT: $\frac{1}{tg^{2}x+(1+tgx)^{2}}+\frac{1}{cot^{2}x+(1+cotx)^{2}}+\frac{(sinx+cosx)^{4}}{4}=\frac{5+sin^{3}2x}{4}$ với $x\in[0,\frac{\pi }{2}]$
nhìn cái đề ko nghĩ đc gì luôn –  conheodat297 31-10-12 09:46 PM
eo nhìn cái đề mà sợ rùi, có anh nào giải dc k bít –  ranganh 25-10-12 10:13 PM
Nhìn khoai thế bạn –  nguyenphuc423 25-10-12 08:17 PM
Đặt $t=\tan \frac{x}{2}, t\ne 0, t\ne \pm 1$ .
thì $\tan x=\frac{2t}{1-t^2}, \sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$ thay vào PT ta thu được
$4t(t-1)(t+1) \left( {t}^{16}+{t}^{15}-6\,{t}^{14}-23\,{t}^{13}+80\,{t}^{12}+37\,{t}^{11}-250\,{t}^{10}+61\,{t}^{9}+606\,{t}^{8}-61\,{t}^{7}-250\,{t}^{6}-37\,{t}^{5}+80\,{t}^{4}+23\,{t}^{3}-6\,{t}^{2}-t+1 \right) =0$
PT bậc $16$ này vô nghiệm nên PT đã cho vô nghiệm.
de thi vo dich my –  nguyenvanton04051997 20-09-13 08:08 PM
Hãy ấn chữ V dưới đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  Trần Nhật Tân 01-11-12 11:37 PM

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