Ta có $(\sqrt{10}+3)(\sqrt{10}-3) = 1 \Rightarrow \sqrt{10}-3 = \dfrac{1}{\sqrt{10}+3}$
Ta có $(\sqrt{10}+3)^{\frac{x-3}{x+1}} < \dfrac{1}{(\sqrt{10}+3)^{\frac{x+1}{x+3}}}$
$\Leftrightarrow (\sqrt{10}+3)^{\frac{x-3}{x+1}+\frac{x+1}{x+3}} <1 =(\sqrt{10}+3)^0$
$\Leftrightarrow \frac{x-3}{x+1}+\frac{x+1}{x+3} <0$ tự xử nốt nhé