9. The solution set of the equation $\sqrt{2 x^{2}+x+3}+\sqrt{x^{2}+x+1}=\sqrt{5 x^{2}+3 x+7}$ is
Let $A=2 x^{2}+x+3, B=x^{2}+x+1$, then $\sqrt{A}+\sqrt{B}=\sqrt{2 A+B}$ $\Rightarrow A+B+2 \sqrt{A B}=2 A+B \Rightarrow 2 \sqrt{A B}=A \Rightarrow A=4 B$ or $A=0$ (discard). Thus, $2 x^{2}+x+3=4 x^{2}+4 x+4 \Rightarrow 2 x^{2}+3 x+1=0 \Rightarrow x=-1,-\frac{1}{2}$. Therefore, the solution set of the original equation is $\left\{-1,-\frac{1}{2}\right\}$.