2, Đk x $\epsilon$ [-1,1] cho nên đặt x=cost , t $\epsilon$ [0; $\pi$]
Khi đó pt trở thành:
$\sqrt{1+\sqrt{1-\cos^{2}t}}$$($$\sqrt{(1+cost)^{3}}$ $-$ $\sqrt{(1-cost)^{3}}$$)$ = $\frac{2}{\sqrt{3}}$ + $\sqrt{\frac{1-cos^{2}t}{3}}$
$\Leftrightarrow$ $\sqrt{1+sint}$ $\left ( \sqrt{8\left ( \frac{1+cost}{2}^{3} \right )}- \sqrt{8\left ( \frac{1-cost}{2}^{3} \right )} \right )$= $\frac{2+sint}{\sqrt{3}}$
$\Leftrightarrow$ 2$\sqrt{2}$$\left ( sin\tfrac{t}{2} +cos\tfrac{t}{2} \right )$$\left ( cos\tfrac{t}{2} -sin\tfrac{t}{2}\right )$$\left ( 1+\frac{sint}{2} \right )$= $\frac{2+sint}{\sqrt{3}}$
Đến đây bạn tự làm nốt sẽ ra được cost = $\frac{1}{\sqrt{6}}$ => x = $\frac{1}{\sqrt{6}}$