Viết lại pt: $4\sqrt{(x+1)(x+2)}-(\sqrt{2}+3)\sqrt{x+1}-3(\sqrt{x+2}-1)=0$$\Leftrightarrow \sqrt{x+1}(\underbrace{4\sqrt{x+2}-3\frac{\sqrt{x+1}}{\sqrt{x+2}+1}-\sqrt{2}-3}_{f(x)})=0$
$+f(x)=0\Leftrightarrow 4x+3(1-\sqrt{x+1})+(1-\sqrt{2})(\sqrt{x+2}-\sqrt{2})=0$
$\Leftrightarrow x(\underbrace{4-\frac{3}{1+\sqrt{x+1}}-\frac{\sqrt{2}-1}{\sqrt{2}+\sqrt{x+2}}}_{g(x)})=0$
$+g(x)=0$ Ta có: $\frac{\sqrt{2}-1}{\sqrt{2}+\sqrt{x+2}}+\frac{3}{1+\sqrt{x+1}}\leq \frac{\sqrt{2}-1}{\sqrt{2}+1}+3<4$
$\Rightarrow g(x)=0$ vô nghiệm