$\frac{10}{(x - 5)(x + 1)} + \frac{x}{x + 1} = \frac{3}{x - 5}$ |*(x − 5)(x + 1)
$\begin{equation*} \begin{cases} 10 + x(x - 5) = 3(x + 1) &\\ x ≠ {-1;5} & \end{cases} \end{equation*}$
$10 + x^2 - 5x = 3x + 3$
$x^2 - 8x + 7 = 0$
(x − 1)(x − 7) = 0
$x_1 = 1$
$x_2 = 7$
$\begin{equation*} \begin{cases} x_1 = 1, x_2 = 7 &\\ x ≠ {-1;5} & \end{cases} \end{equation*}$
Ответ:
$x_1 = 1$
$x_2 = 7$

$\frac{17}{(x - 3)(x+ 4)} - \frac{1}{x - 3} = \frac{x}{x + 4}$ |*(x − 3)(x + 4)
$\begin{equation*} \begin{cases} 17 - (x + 4) = x(x - 3) &\\ x ≠ {-4;3} & \end{cases} \end{equation*}$
$17 - x - 4 = x^2 - 3x$
$x^2 - 2x - 13 = 0$
D = 1 + 13 = 14
$x = ±\sqrt{14}$
$\begin{equation*} \begin{cases} x = ±\sqrt{14} &\\ x ≠ {-4;3} & \end{cases} \end{equation*}$
Ответ: $x = ±\sqrt{14}$

$\frac{4}{(x + 1)^2} - \frac{1}{(x - 1)^2} + \frac{1}{x^2 - 1} = 0$
$\begin{equation*} \begin{cases} 4(x - 1)^2 - (x + 1)^2 + x^2 - 1 = 0 &\\ x ≠ ±1 & \end{cases} \end{equation*}$
$4x^2 - 8x + 4 - x^2 - 2x - 1 + x^2 - 1 = 0$
$4x^2 - 10x + 2 = 0$
$2x^2 - 5x + 1 = 0$
D = 25 − 4 * 2 = 17
$x = \frac{5 ± \sqrt{17}}{4}$
$\begin{equation*} \begin{cases} x = \frac{5 ± \sqrt{17}}{4} &\\ x ≠ ±1 & \end{cases} \end{equation*}$
Ответ: $x = \frac{5 ± \sqrt{17}}{4}$

$\frac{4}{9x^2 - 1} + \frac{1}{3x^2 - x} = \frac{4}{9x^2 - 6x + 1} |*x(3x - 1)^2(3x + 1)$
$\begin{equation*} \begin{cases} 4x(3x - 1) + (3x - 1)(3x + 1) = 4x(3x + 1) &\\ x ≠ {0;±\frac{1}{3}} & \end{cases} \end{equation*}$
$12x^2 - 4x + 9x^2 - 1 = 12x^2 + 4x$
$9x^2 - 8x - 1 = 0$
(9x + 1)(x − 1) = 0
$x_1 = -\frac{1}{9}$
$x_2 = 1$
$\begin{equation*} \begin{cases} x_1 = -\frac{1}{9}, x_2 = 1 &\\ x ≠ {0;±\frac{1}{3}} & \end{cases} \end{equation*}$
Ответ:
$x_1 = -\frac{1}{9}$;
$x_2 = 1$.