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

$$
2^{x^{2}+2 x-4 x^{4}}=\frac{x^{2}}{x+1}
$$
1. For the right side to be defined, it must hold that $x \neq -1$. The given equation reduces to $2^{x^{2}+2 x}(x+1)=2^{4 x^{4}} x^{2}$, and multiplying both sides by 2, we observe that this is equivalent to $2^{(x+1)^{2}}(x+1)=2^{\left(2 x^{2}\right)^{2}}\left(2 x^{2}\right)$. Notice that the function $f: t \mapsto 2^{t^{2}} t$ is strictly increasing (indeed, we have $f^{\prime}(t)=2^{t^{2}} \ln 2 \cdot 2 t \cdot t + 2^{t^{2}} = 2^{t^{2}+1} t^{2} \ln 2 + 2^{t^{2}} > 0$), and thus injective. Therefore, it must hold that $x+1=2 x^{2}$, and the solutions to the given equation are $x_{1/2}=\frac{1 \pm \sqrt{1^{2}+4 \cdot 2}}{4}=\frac{1 \pm 3}{4}$, i.e., $x_{1}=1$ and $x_{2}=-\frac{1}{2}$ (both satisfy the condition $x \neq -1$).