1. In the set of real numbers, solve the equation:

$$
\frac{5 x}{x^{2}+3 x+6}+\frac{7 x}{x^{2}+7 x+6}=1
$$
1. Let's make the substitution $x^{2}+3 x+6=t$. The equation reduces to $\frac{5 x}{t}+\frac{7 x}{t+4 x}=1$, i.e., $5 x t+20 x^{2}+7 x t=t^{2}+4 t x$ (with the conditions $t \neq 0$ and $t+4 x \neq 0$), which is equivalent to $t^{2}-8 t x-20 x^{2}=0$. Dividing both sides by $x^{2}$ (with the condition $x^{2} \neq 0$) the equation further reduces to $\left(\frac{t}{x}\right)^{2}-8 \frac{t}{x}-20=0$, and by introducing the new substitution $k=\frac{t}{x}$ we get $k^{2}-8 k-20=0$. The solutions to this quadratic equation are $k_{1 / 2}=\frac{8 \pm \sqrt{8^{2}-4 \cdot(-20)}}{2}=\frac{8 \pm \sqrt{144}}{2}=\frac{8 \pm 12}{2}$, i.e., $k_{1}=\frac{8+12}{2}=10$ and $k_{2}=\frac{8-12}{2}=-2$. For $k=10$, after returning the substitution, we have $x^{2}+3 x+6=t=k x=10 x$, i.e., $x^{2}-7 x+6=0$, whose solution gives $x_{1 / 2}=\frac{7 \pm \sqrt{(-7)^{2}-4 \cdot 6}}{2}=\frac{7 \pm 5}{2}$, i.e., $x_{1}=6$ and $x_{2}=1$. For $k=-2$, after returning the substitution, we have $x^{2}+3 x+6=t=k x=-2 x$, i.e., $x^{2}+5 x+6=0$, whose solution gives $x_{3 / 4}=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 6}}{2}=\frac{-5 \pm 1}{2}$, i.e., $x_{3}=-2$ and $x_{4}=-3$. All four obtained solutions satisfy the conditions of definition, so they are indeed the solutions to the original equation.