Giải phương trình.

Câu $2:$

$x+\dfrac{x}{\sqrt{x^2-1}}=2\sqrt{2}$

Điều kiện: $x^2-1>0\Rightarrow|x|>1$

$1+\dfrac{1}{\sqrt{x^2-1}}=\dfrac{2\sqrt{2}}{x}$

Đặt $x=\sec\theta\Rightarrow\sqrt{x^2-1}=\tan\theta,x>1\Rightarrow\theta\in\left(0,\dfrac{\pi}{2}\right)$

$\Rightarrow\sec\theta+\dfrac{\sec\theta}{\tan\theta}=2\sqrt{2}$

$\Leftrightarrow\dfrac{1}{\cos\theta}+\dfrac{1}{\sin\theta}=2\sqrt{2}$

$\Rightarrow\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=2\sqrt{2}$

Đặt $\sin\theta+\cos\theta=t\Rightarrow t\in(0,\sqrt{2})$ và $\sin\theta\cos\theta=\dfrac{t^2-1}{2}$

$\Rightarrow\dfrac{t}{\dfrac{t^2-1}{2}}=2\sqrt{2}\Rightarrow\dfrac{2t}{t^2-1}=2\sqrt{2}$

$\Leftrightarrow\dfrac{t}{t^2-1}=\sqrt{2}$

$\Rightarrow t=\sqrt{2}(t^2-1)\Rightarrow\sqrt{2}t^2-t-\sqrt{2}=0$

$\Rightarrow t=\dfrac{1\pm3}{2\sqrt{2}}$

$\Rightarrow\begin{cases}t_1=\dfrac{1+3}{2\sqrt{2}}=\sqrt{2} \\ t_2=\dfrac{1-3}{2\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\end{cases}$

Vì $t=\sin\theta+\cos\theta>0\Rightarrow t=\sqrt{2}$

$\Rightarrow\sin\theta+\cos\theta=\sqrt{2}\Rightarrow\theta=\dfrac{\pi}{4}$

$\Rightarrow x=\sec\theta=\sec\dfrac{\pi}{4}=\sqrt{2}$