$$8\sqrt 2 \sin x.\cos 2x + 1 = \tan 4x + \tan x + \tan x.\tan 4x$$
MOD: Chú ý tiêu đề bạn nhé.
Edited by Mai Duc Khai, 25-05-2013 - 20:46.
$$8\sqrt 2 \sin x.\cos 2x + 1 = \tan 4x + \tan x + \tan x.\tan 4x$$
MOD: Chú ý tiêu đề bạn nhé.
Edited by Mai Duc Khai, 25-05-2013 - 20:46.
ĐK:$\left\{\begin{matrix}cosx\neq 0 &\\ cos4x\neq 0& \end{matrix}\right.$
pt:$8\sqrt{2}sinx.cos2x+(1-tanx.tan4x)=tan4x+tanx$
$\Leftrightarrow 8\sqrt{2}sinx.cos2x+\frac{cos5x}{cosx.cos4x}=\frac{sin5x}{cosx.cos4x}$
0 members, 1 guests, 0 anonymous users