ĐK: $x^2-1 \leq 0$ và $3-x \leq 0$
PT $\iff (x+1)-\sqrt{x^2-1}+(x+3)-\sqrt{3-x}=0$
$\iff \dfrac{2x+2}{x+1+\sqrt{x^2-1}}+\dfrac{x^2+7x+6}{x+3+\sqrt{3-x}}=0$
$\iff (x+1)(\dfrac{2}{x+1+\sqrt{x^2-1}}+\dfrac{x+6}{x+3+\sqrt{3-x}})=0$
$\iff x=-1$ (vì $\dfrac{2}{x+1+\sqrt{x^2-1}}+\dfrac{x+6}{x+3+\sqrt{3-x}}>0$)
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