$\frac{2x + 1}{2x - 1} - \frac{3(2x - 1)}{7(2x + 1)} + \frac{8}{1 - 4x^2} = 0$
$\frac{2x + 1}{2x - 1} - \frac{3(2x - 1)}{7(2x + 1)} - \frac{8}{4x^2 - 1} = 0$|*7(2x + 1)(2x − 1)
$\begin{equation*} \begin{cases} 7(2x + 1)^2 - 3(2x - 1)^2 - 7 * 8 = 0 &\\ x ≠ ±\frac{1}{2} & \end{cases} \end{equation*}$
$7(4x^2 + 4x + 1) - 3(4x^2 - 4x + 1) - 56 = 0$
$16x^2 + 40x - 52 = 0$|:4
$4x^2 + 10x - 13 = 0$
$D = 5^2 + 4 * 13 = 77$
$\begin{equation*} \begin{cases} x = \frac{-5 ± \sqrt{77}}{4} &\\ x ≠ ±\frac{1}{2} & \end{cases} \end{equation*}$
Ответ: $x_{1,2} = \frac{-5 ± \sqrt{77}}{4}$

$\frac{y}{y^2 - 9} - \frac{1}{y^2 + 3y} + \frac{3}{6y + 2y^2} = 0$
$\frac{y}{(y - 3)(y + 3)} - \frac{1}{y(y + 3)} + \frac{3}{2y(y + 3)} = 0$|*2y(y + 3)(y − 3)
$\begin{equation*} \begin{cases} 2y^2 - 2(y - 3) + 3(y - 3) = 0 &\\ y ≠ ±3 & \end{cases} \end{equation*}$
$2y^2 + y - 3 = 0$
(2y + 3)(y − 1) = 0
$\begin{equation*} \begin{cases} y_1 = -1,5, y_2 = 1 &\\ y ≠ ±3 & \end{cases} \end{equation*}$
Ответ:
$y_1 = -1,5$;
$y_2 = 1$.

$\frac{2y - 1}{14y^2 + 7y} + \frac{8}{12y^2 - 3} = \frac{2y + 1}{6y^2 - 3y}$
$\frac{2y - 1}{7y(2y + 1)} + \frac{8}{3(4y^2 - 1)} = \frac{2y + 1}{3y(2y - 1)}$|*21y(2y − 1)(2y + 1)
$\begin{equation*} \begin{cases} 3(2y - 1)^2 + 8 * 7y = 7(2y + 1)^2 &\\ y ≠ ±\frac{1}{2} & \end{cases} \end{equation*}$
$3(4y^2 - 4y + 1) + 56y = 7(4y^2 + 4y + 1)$
$16y^2 - 16y + 4 = 0$
$4y^2 - 4y + 1 = 0$
$(2y - 1)^2 = 0$
$\begin{equation*} \begin{cases} y = \frac{1}{2} &\\ y ≠ ±\frac{1}{2} & \end{cases} \end{equation*}$
Ответ: y = ∅ − пустое множество, нет решений.

$\frac{3}{x^2 - 9} - \frac{1}{9 - 6x + x^2} = \frac{3}{2x^2 + 6x}$
$\frac{3}{(x - 3)(x + 3)} - \frac{1}{(x - 3)^2} = \frac{3}{2x(x + 3)} |*2x(x + 3)(x - 3)^2$
$\begin{equation*} \begin{cases} 3 * 2x(x - 3) - 2x(x + 3) = 3(x - 3)^2 &\\ x ≠ ±3 & \end{cases} \end{equation*}$
$6x^2 - 18x - 2x^2 - 6x = 3(x^2 - 6x + 9)$
$x^2 - 6x - 27 = 0$
(x + 3)(x − 9) = 0
$\begin{equation*} \begin{cases} x_1 = -3, x_2 = 9 &\\ x ≠ ±3 & \end{cases} \end{equation*}$
Ответ: x = 9

$\frac{9x + 12}{x^3 - 64} - \frac{1}{x^2 + 4x + 16} = \frac{1}{x - 4}$
$\frac{9x + 12}{(x - 4)(x^2 + 4x + 16)} - \frac{1}{x^2 + 4x + 16} = \frac{1}{x - 4} |*(x - 4)(x^2 + 4x + 16)$
$\begin{equation*} \begin{cases} (9x + 12) - (x - 4) = x^2 + 4x + 16 &\\ x ≠ 4 & \end{cases} \end{equation*}$
$9x + 12 - x + 4 - x^2 - 4x - 16 = 0$
$x^2 - 4x = 0$
x(x − 4) = 0
$\begin{equation*} \begin{cases} x_1 = 0, x_2 = 4 &\\ x ≠ 4 & \end{cases} \end{equation*}$
Ответ: x = 0

$\frac{3}{8y^3 + 1} - \frac{1}{2y + 1} = \frac{y + 3}{4y^2 - 2y + 1}$
$\frac{3}{(2y + 1)(4y^2 - 2y + 1)} - \frac{1}{2y + 1} = \frac{y + 3}{4y^2 - 2y + 1} |*(2y + 1)(4y^2 - 2y + 1)$
$\begin{equation*} \begin{cases} 3 - (4y^2 - 2y + 1) = (y + 3)(2y + 1) &\\ y ≠ -\frac{1}{2} & \end{cases} \end{equation*}$
$3 - 4y^2 + 2y - 1 = 2y^2 + 7y + 3$
$-6y^2 - 5y - 1 = 0$
$6y^2 + 5y + 1 = 0$
(3y + 1)(2y + 1) = 0
$\begin{equation*} \begin{cases} y_1 = -\frac{1}{3}, y_2 = -\frac{1}{2} &\\ y ≠ -\frac{1}{2} & \end{cases} \end{equation*}$
Ответ: $y = -\frac{1}{3}$

$\frac{32}{x^3 - 2x^2 - x + 2} + \frac{1}{(x - 1)(x - 2)} = \frac{1}{x + 1}$
$x^3 - 2x^2 - x + 2 = x^2(x - 2) - (x - 2) = (x^2 - 1)(x - 2)$
$\frac{32}{(x - 1)(x + 1)(x - 2)} + \frac{1}{(x - 1)(x - 2)} = \frac{1}{x + 1}$|*(x − 1)(x + 1)(x − 2)
$\begin{equation*} \begin{cases} 32 + x + 1 = (x - 1)(x - 2) &\\ x ≠ {±1;2} & \end{cases} \end{equation*}$
$32 + x + 1 = x^2 - x - 2x + 2$
$x^2 - 4x - 31 = 0$
$D = 2^2 + 31 = 4 + 31 = 35$
$\begin{equation*} \begin{cases} x = 2 ± \sqrt{35} &\\ x ≠ {±1;2} & \end{cases} \end{equation*}$
Ответ: $x_{1,2} = 2 ± \sqrt{35}$

$\frac{1}{3(x - 4)} + \frac{1}{2(x^2 + 3)} + \frac{1}{x^3 - 4x^2 + 3x - 12} = 0$
$x^3 - 4x^2 + 3x - 12 = x^2(x - 4) + 3(x - 4) = (x^2 + 3)(x - 4)$
$\frac{1}{3(x - 4)} + \frac{1}{2(x^2 + 3)} + \frac{1}{(x^2 + 3)(x - 4)} = 0 |*6(x^2 + 3)(x - 4)$
$\begin{equation*} \begin{cases} 2(x^2 + 3) + 3(x - 4) + 6 = 0 &\\ x ≠ 4 & \end{cases} \end{equation*}$
$2x^2 + 6 + 3x - 12 + 6 = 0$
$2x^2 + 3x = 0$
x(2x + 3) = 0
$\begin{equation*} \begin{cases} x_1 = 0, x_2 = -1,5 &\\ x ≠ 4 & \end{cases} \end{equation*}$
Ответ:
$x_1 = 0$;
$x_2 = -1,5$.