1. Solve the equation

$$
\sqrt{x^{2}+x}+\sqrt{1+\frac{1}{x^{2}}}=\sqrt{x+3}
$$
Solution. From the given equation by squaring follows the equation

$$
\left(x-\frac{1}{x}\right)^{2}+2 \sqrt{\left(x^{2}+x\right)\left(1+\frac{1}{x^{2}}\right)}=0
$$

From this, it follows that $x-\frac{1}{x}=0$, i.e., $x=1$ or $x=-1$. By direct verification, we find that $x=1$ is not a solution to the original equation, while $x=-1$ is its solution.