1. Solve the equation

$$
\sqrt{x^{2}+x}+\sqrt{1+\frac{1}{x^{2}}}=\sqrt{x+3}
$$

in the set of real numbers.
Solution. The domain of the equation is

$$
D=\left\{x \mid x^{2}+x \geq 0, x \neq 0, x \geq-3\right\}=[-3,-1] \cup(0,+\infty)
$$

By framing the equation, we get:

$$
x^{2}+x+1+\frac{1}{x^{2}}+2 \sqrt{\left(x^{2}+x\right)\left(1+\frac{1}{x^{2}}\right)}=x+3
$$

from which we have

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

or equivalently,

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

The last equation is equivalent to the system

$$
\left\{\begin{array}{l}
\left(x-\frac{1}{x}\right)^{2}=0 \\
\sqrt{(x+1)\left(x+\frac{1}{x}\right)}=0
\end{array}\right.
$$

We have

$$
\left\{\begin{array}{l}
x-\frac{1}{x}=0 \\
(x+1)\left(x+\frac{1}{x}\right)=0
\end{array}\right.
$$

i.e.,

$$
\left\{\begin{array}{l}
x-\frac{1}{x}=0 \\
x=-1
\end{array}\right.
$$

Since $x=-1 \in D$ and satisfies $x-\frac{1}{x}=0$, $x=-1$ is the solution to the equation.