$\begin{array}{l}
\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, x \neq-1,-2,-4 \\
\Rightarrow \frac{x+2+2 x+2}{(x+1)(x+2)}=\frac{5}{x+4} \\
\Rightarrow \frac{3 x+4}{x^{2}+3 x+2}=\frac{5}{x+4} \\
\Rightarrow(3 x+4)(x+4)=5\left(x^{2}+3 x+2\right) \\
\Rightarrow 3 x^{2}+16 x+16=5 x^{2}+15 x+10 \\
\Rightarrow 2 x^{2}-x-6=0 \\
\Rightarrow 2 x^{2}-4 x+3 x-6=0 \\
\Rightarrow 2 x(x-2)+3(x-2)=0 \\
\Rightarrow(x-2)(2 x+3)=0
\end{array}$
$\begin{array}{l}
\Rightarrow 3 x^{2}+16 x+16=5 x^{2}+15 x+10 \\
\Rightarrow 2 x^{2}-x-6=0 \\
\Rightarrow 2 x^{2}-4 x+3 x-6=0 \\
\Rightarrow 2 x(x-2)+3(x-2)=0 \\
\Rightarrow(x-2)(2 x+3)=0 \\
\Rightarrow x-2=0 \text { or } 2 x+3=0 \\
\Rightarrow x=2 \text { or } x=-\frac{3}{2}
\end{array}$

Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.