$\frac{x\sqrt{3} + \sqrt{2}}{x\sqrt{3} - \sqrt{2}} + \frac{x\sqrt{3} - \sqrt{2}}{x\sqrt{3} + \sqrt{2}} = \frac{10x}{3x^2 - 2}|*(x\sqrt{3} - \sqrt{2})(x\sqrt{3} + \sqrt{2})$
$\begin{equation*} \begin{cases} (x\sqrt{3} + \sqrt{2})^2 + (x\sqrt{3} - \sqrt{2})^2 = 10x &\\ x ≠ ±\sqrt{\frac{2}{3}} & \end{cases} \end{equation*}$
$3x^2 + 2\sqrt{6}x + 2 + 3x^2 - 2\sqrt{6}x + 2 = 10x$
$6x^2 - 10x + 4 = 0$
$3x^2 - 5x + 2 = 0$
(3x − 2)(x − 1)
$\begin{equation*} \begin{cases} x_1 = \frac{2}{3}, x_2 = 1 &\\ x ≠ ±\sqrt{\frac{2}{3}} & \end{cases} \end{equation*}$
Ответ:
$x_1 = \frac{2}{3}$;
$x_2 = 1$.

$\frac{1 - y\sqrt{5}}{1 + \sqrt{5}} + \frac{1 + y\sqrt{5}}{1 - y\sqrt{5}} = \frac{9y}{1 - 5y^2} |*(1 + y\sqrt{5})(1 - y\sqrt{5})$
$\begin{equation*} \begin{cases} (1 - y\sqrt{5})^2 + (1 + y\sqrt{5})^2 = 9y &\\ y ≠ ±\frac{1}{\sqrt{5}} & \end{cases} \end{equation*}$
$1 - 2y\sqrt{5} + 5y^2 + 1 + 2y\sqrt{5} + 5y^2 = 9y$
$10y^2 - 9y + 2 = 0$
(5x − 2)(2x − 1) = 0
$\begin{equation*} \begin{cases} y_1 = \frac{2}{5}, y_2 = \frac{1}{2} &\\ y ≠ ±\frac{1}{\sqrt{5}} & \end{cases} \end{equation*}$
Ответ:
$y_1 = \frac{2}{5}$;
$y_2 = \frac{1}{2}$.