1. Solve the equation $\sqrt{x^{2}+x}+\sqrt{1+\frac{1}{x^{2}}}=\sqrt{x+3}$.
Solution. For the domain, we get $D=[-3,-1] \cup(0,+\infty)$. Then,

$$
\begin{aligned}
& x^{2}+x+1+\frac{1}{x^{2}}+2 \sqrt{x^{2}+x} \sqrt{1+\frac{1}{x^{2}}}=x+3 \\
& x^{2}-2+\frac{1}{x^{2}}+2 \sqrt{(x+1) \frac{x^{2}+1}{x}}=0 \\
& \left(x-\frac{1}{x}\right)^{2}+2 \sqrt{(x+1) \frac{x^{2}+1}{x}}=0
\end{aligned}
$$

From this,

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

and we get that $x=-1$, which is a solution to the given equation.