haibinh232

$8cosx=\frac{\sqrt{3}}{cosx}+sinx$       ĐK: $cosx\neq 0$

$\Leftrightarrow 8cos^{2}x-sinx-\sqrt{3}=0$

$\Leftrightarrow8sin^{2}x+sinx+(\sqrt{3}-8)=0$

$\Leftrightarrow \left[\begin{matrix}sinx=\frac{-1+\sqrt{257-32\sqrt{3}}}{32}\simeq 0,412\\sinx=\frac{-1-\sqrt{257-32\sqrt{3}}}{32}\simeq -0,475 \end{matrix}\right.$

Nếu bạn đủ kiên nhẫn thì làm tiếp   

P  $=\sqrt{\frac{5-2\sqrt{6}}{5+\sqrt{6}}}+\sqrt{\frac{5+2\sqrt{6}}{5-\sqrt{6}}}$

    $=\sqrt{\frac{(5-2\sqrt{6})(5-\sqrt{6})}{(5+\sqrt{6})(5-\sqrt{6})}}+\sqrt{\frac{(5+2\sqrt{6})(5+\sqrt{6})}{(5+\sqrt{6})(5-\sqrt{6})}}$

    $=\sqrt{\frac{37-15\sqrt{6}}{19}}+\sqrt{\frac{37+15\sqrt{6}}{19}}$

$\Leftrightarrow P^{2}=\frac{37-15\sqrt{6}}{19}+\frac{37+15\sqrt{6}}{19}+2\sqrt{\frac{(37-15\sqrt{6})(37+15\sqrt{6})}{19^{2}}}$

$\Leftrightarrow P^{2}=\frac{74}{19}+2\sqrt{\frac{1}{19}}$

$\Leftrightarrow P^{2} = \frac{74+2\sqrt{19}}{19}$

$\Leftrightarrow P = \sqrt{\frac{74+2\sqrt{19}}{19}}\simeq 2.0865$

Nếu đề là $2x^{2}+8x-3\sqrt{x^{2}+4x-5}=12$

$\Leftrightarrow 2(x^{2}+4x-6)-3$

Đặt $t=\sqrt{x^{2}+4x-5}$, $t\geqslant 0$

      $\Leftrightarrow t^{2}-1=x^{2}+4x-6$

pt $\Leftrightarrow 2(t^{2}-1)+t=0$

$\Leftrightarrow 2t^{2}-3t-2=0$

$\Leftrightarrow \left[\begin{matrix}t=-1\\t=4 \end{matrix}\right.$

$\Leftrightarrow t=4$

$\Leftrightarrow \sqrt{x^{2}+4x-5}=4$

$\Leftrightarrow x^{2}+4x-21=0$

$\Leftrightarrow \left[\begin{matrix}x=3\\x=-7\end{matrix}\right.$

$x^{2}+(\sqrt{x}+1)(5x-6\sqrt{x}-6)=0$

$\Leftrightarrow x^{2}+(\sqrt{x}+1)[5x-6(\sqrt{x}+1)]=0$

$\Leftrightarrow x^{2}+5(\sqrt{x}+1)x-6(\sqrt{x}+1)^{2}=0$

$\Delta = 25(\sqrt{x}+1)^{2}+24(\sqrt{x}+1)^{2}=49(\sqrt{x}+1)^{2}$

$\Rightarrow \sqrt{\Delta }= 7(\sqrt{x}+1)$

$\Rightarrow \left[\begin{matrix}x=\sqrt{x}+1\\x=-6(\sqrt{x}+1) \end{matrix}\right.$

$\Rightarrow \left[\begin{matrix}x-\sqrt{x}-1=0\\x+6\sqrt{x}+6=0 \end{matrix}\right.$

$\Leftrightarrow \sqrt{x}=\frac{1+\sqrt{5}}{2}\Leftrightarrow x=\frac{3+\sqrt{5}}{2}$ là nghiệm duy nhất

Bài 1

$\frac{1 + sin^{4}\alpha - cos^{4}\alpha }{1-sin^{6}\alpha -cos^{6}\alpha }=\frac{2}{3cos^{2}\alpha }$

$\Leftrightarrow \frac{1+(sin^{2}\alpha +cos^{2}\alpha)(sin^{2}\alpha -cos^{2}\alpha) }{1-(sin^{2}\alpha +cos^{2}\alpha )(sin^{4}\alpha +cos^{4}\alpha -sin^{2}\alpha. cos^{2}\alpha )}=\frac{2}{3cos^{2}\alpha }$

$\Leftrightarrow\frac{1-cos2\alpha }{1-[(sin^{2}\alpha +cos^{2}\alpha )^{2}-3sin^{2}\alpha .cos^{2}\alpha ]}=\frac{2}{3cos^{2}\alpha }$

$\Leftrightarrow\frac{2sin^{2}\alpha }{3sin^{2}\alpha .cos^{2}\alpha }=\frac{2}{3cos^{2}\alpha }$

$\Leftrightarrow\frac{2}{3cos^{2}\alpha }=\frac{2}{3cos^{2}\alpha }$ (Đúng)