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

$$
\frac{1}{x^{2}}+\frac{1}{(\sqrt{5}-x)^{2}}=7
$$
Solution. Clearly, $x \notin\{0, \sqrt{5}\}$. We introduce the substitution $y=\sqrt{5}-x$ and obtain the system of equations

$$
\left\{\begin{array}{l}
x+y=\sqrt{5} \\
\frac{1}{x^{2}}+\frac{1}{y^{2}}=7
\end{array}\right.
$$

We have,

$$
7=\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{(x+y)^{2}-2 x y}{(x y)^{2}}=\frac{5-2 x y}{(x y)^{2}}
$$

hence $7(x y)^{2}+2 x y-5=0$, from which we get $x y=\frac{5}{7}$ or $x y=-1$. From $x+y=\sqrt{5}$ and $x y=\frac{5}{7}$, it follows that $x$ and $y$ are solutions to the quadratic equation $z^{2}-z \sqrt{5}+\frac{5}{7}=0$, that is, $x, y=\frac{7 \sqrt{5} \pm \sqrt{105}}{14}$. From $x+y=\sqrt{5}$, we get $x=\sqrt{5}-y=\sqrt{5}-\frac{7 \sqrt{5} \pm \sqrt{105}}{14}=\frac{7 \sqrt{5} \mp \sqrt{105}}{14}$, so the equation in $x$ has the solutions $x_{1,2}=\frac{7 \sqrt{5} \pm \sqrt{105}}{14}$. From $x+y=\sqrt{5}$ and $x y=-1$, it follows that $x$ and $y$ are solutions to the quadratic equation $z^{2}-z \sqrt{5}+-1=0$, that is, $x, y=\frac{\sqrt{5} \pm 3}{2}$. From $x+y=\sqrt{5}$, we get $x=\sqrt{5}-y=\sqrt{5}-\frac{\sqrt{5} \pm 3}{2}=\frac{\sqrt{5} \mp 3}{2}$, so the equation in $x$ has the solutions $x_{3,4}=\frac{\sqrt{5} \pm 3}{2}$.