$\frac{5}{x^2 - 4} + \frac{2x}{x + 2} = 2$
$\frac{5}{(x - 2)(x + 2)} + \frac{2x}{x + 2} - 2 = 0$
$\frac{5 + 2x(x - 2) - 2(x^2 - 4)}{(x - 2)(x + 2)} = 0$
$\frac{2x^2 - 4x + 5 - 2x^2 + 8}{(x - 2)(x + 2)} = 0$
$\frac{13 - 4x}{(x - 2)(x + 2)} = 0$
$\begin{equation*} \begin{cases} x - 2 ≠ 0 &\\ x + 2 ≠ 0 &\\ 13 - 4x = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 2 &\\ x ≠ -2 &\\ 4x = 13 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 2 &\\ x ≠ -2 &\\ x = \frac{13}{4} = 3\frac{1}{4} & \end{cases} \end{equation*}$
Ответ: $x = 3\frac{1}{4}$

$\frac{2}{6x + 1} + \frac{3}{6x - 1} = \frac{30x + 9}{36x^2 - 1}$
$\frac{2}{6x + 1} + \frac{3}{6x - 1} - \frac{30x + 9}{(6x - 1)(6x + 1)} = 0$
$\frac{2(6x - 1) + 3(6x + 1) - (30x + 9)}{(6x - 1)(6x + 1)} = 0$
$\frac{12x - 2 + 18x + 3 - 30x - 9}{(6x - 1)(6x + 1)} = 0$
$\frac{-8}{(6x - 1)(6x + 1)} = 0$
−8 ≠ 0
Ответ: нет корней

$\frac{6x + 14}{x^2 - 9} + \frac{7}{x^2 + 3x} = \frac{6}{x - 3}$
$\frac{6x + 14}{(x - 3)(x + 3)} + \frac{7}{x(x + 3)} - \frac{6}{x - 3} = 0$
$\frac{x(6x + 14) + 7(x - 3) - 6x(x + 3)}{x(x - 3)(x + 3)} = 0$
$\frac{6x^2 + 14x + 7x - 21 - 6x^2 - 18x}{x(x - 3)(x + 3)} = 0$
$\frac{3x - 21}{x(x - 3)(x + 3)} = 0$
$\begin{equation*} \begin{cases} x ≠ 0 &\\ x - 3 ≠ 0 &\\ x + 3 ≠ 0 &\\ 3x - 21 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 0 &\\ x ≠ 3 &\\ x ≠ -3 &\\ 3x = 21 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 0 &\\ x ≠ 3 &\\ x ≠ -3 &\\ x = 7 & \end{cases} \end{equation*}$
Ответ: x = 7

$\frac{2y^2 + 5}{1 - y^2} + \frac{y + 1}{y - 1} = \frac{4}{y + 1}$
$\frac{-(2y^2 + 5)}{y^2 - 1} + \frac{y + 1}{y - 1} - \frac{4}{y + 1} = 0$
$\frac{-2y^2 - 5}{(y - 1)(y + 1)} + \frac{y + 1}{y - 1} - \frac{4}{y + 1} = 0$
$\frac{-2y^2 - 5 + (y + 1)^2 - 4(y - 1)}{(y - 1)(y + 1)} = 0$
$\frac{-2y^2 - 5 + y^2 + 2y + 1 - 4y + 4}{(y - 1)(y + 1)} = 0$
$\frac{-y^2 - 2y}{(y - 1)(y + 1)} = 0$
$\frac{-(y^2 + 2y)}{(y - 1)(y + 1)} = 0$
$\frac{-y(y + 2)}{(y - 1)(y + 1)} = 0$
$\begin{equation*} \begin{cases} y - 1 ≠ 0 &\\ y + 1 ≠ 0 &\\ -y = 0 &\\ y + 2 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y ≠ 1 &\\ y ≠ -1 &\\ y = 0 &\\ y = -2 & \end{cases} \end{equation*}$
Ответ: y = −2, y = 0.

$\frac{2x - 1}{2x + 1} = \frac{2x + 1}{2x - 1} + \frac{4}{1 - 4x^2}$
$\frac{2x - 1}{2x + 1} - \frac{2x + 1}{2x - 1} - \frac{4}{1 - 4x^2} = 0$
$\frac{2x - 1}{2x + 1} - \frac{2x + 1}{2x - 1} + \frac{4}{4x^2 - 1} = 0$
$\frac{2x - 1}{2x + 1} - \frac{2x + 1}{2x - 1} + \frac{4}{(2x - 1)(2x + 1)} = 0$
$\frac{(2x - 1)^2 - (2x + 1)^2 + 4}{(2x - 1)(2x + 1)} = 0$
$\frac{4x^2 - 4x + 1 - (4x^2 + 4x + 1) + 4}{(2x - 1)(2x + 1)} = 0$
$\frac{4x^2 - 4x + 1 - 4x^2 - 4x - 1 + 4}{(2x - 1)(2x + 1)} = 0$
$\frac{-8x + 4}{(2x - 1)(2x + 1)} = 0$
$\begin{equation*} \begin{cases} 2x - 1 ≠ 0 &\\ 2x + 1 ≠ 0 &\\ -8x + 4 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 2x ≠ 1 &\\ 2x ≠ -1 &\\ -8x = -4 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 0,5 &\\ x ≠ -0,5 &\\ x = 0,5 & \end{cases} \end{equation*}$
Ответ: нет корней

$\frac{7}{(x + 2)(x - 3)} - \frac{4}{(x - 3)^2} = \frac{3}{(x + 2)^2}$
$\frac{7}{(x + 2)(x - 3)} - \frac{4}{(x - 3)^2} - \frac{3}{(x + 2)^2} = 0$
$\frac{7(x + 2)(x - 3) - 4(x + 2)^2 - 3(x - 3)^2}{(x + 2)^2(x - 3)^2} = 0$
$\frac{7(x^2 + 2x - 3x - 6) - 4(x^2 + 4x + 4) - 3(x^2 - 6x + 9)}{(x + 2)^2(x - 3)^2} = 0$
$\frac{7(x^2 - x - 6) - 4(x^2 + 4x + 4) - 3(x^2 - 6x + 9)}{(x + 2)^2(x - 3)^2} = 0$
$\frac{7x^2 - 7x - 42 - 4x^2 - 16x - 16 - 3x^2 + 18x - 27}{(x + 2)^2(x - 3)^2} = 0$
$\frac{-5x - 85}{(x + 2)^2(x - 3)^2} = 0$
$\begin{equation*} \begin{cases} x + 2 ≠ 0 &\\ x - 3 ≠ 0 &\\ -5x - 85 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ -2 &\\ x ≠ 3 &\\ -5x = 85 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ -2 &\\ x ≠ 3 &\\ x = -17 & \end{cases} \end{equation*}$
Ответ: x = −17

$\frac{2x - 1}{x + 4} - \frac{3x - 1}{4 - x} = \frac{6x + 64}{x^2 - 16} + 4$
$\frac{2x - 1}{x + 4} + \frac{3x - 1}{x - 4} - \frac{6x + 64}{(x - 4)(x + 4)} - 4 = 0$
$\frac{(2x - 1)(x - 4) + (3x - 1)(x + 4) - (6x + 64) - 4(x^2 - 16)}{(x - 4)(x + 4)} = 0$
$\frac{2x^2 - x - 8x + 4 + 3x^2 - x + 12x - 4 - 6x - 64 - 4x^2 + 64}{(x - 4)(x + 4)} = 0$
$\frac{x^2 - 4x}{(x - 4)(x + 4)} = 0$
$\frac{x(x - 4)}{(x - 4)(x + 4)} = 0$
$\begin{equation*} \begin{cases} x - 4 ≠ 0 &\\ x + 4 ≠ 0 &\\ x = 0 &\\ x - 4 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x ≠ 4 &\\ x ≠ -4 &\\ x = 0 &\\ x = 4 & \end{cases} \end{equation*}$
Ответ: x = 0

$\frac{2x - 6}{x^2 - 36} - \frac{x - 3}{x^2 - 6x} - \frac{x - 1}{x^2 + 6x} = 0$
$\frac{2x - 6}{(x - 6)(x + 6)} - \frac{x - 3}{x(x - 6)} - \frac{x - 1}{x(x + 6)} = 0$
$\frac{x(2x - 6) - (x - 3)(x + 6) - (x - 1)(x - 6)}{x(x - 6)(x + 6)} = 0$
$\frac{2x^2 - 6x - (x^2 - 3x + 6x - 18) - (x^2 - x - 6x + 6)}{x(x - 6)(x + 6)} = 0$
$\frac{2x^2 - 6x - x^2 + 3x - 6x + 18 - x^2 + x + 6x - 6}{x(x - 6)(x + 6)} = 0$
$\frac{-2x + 12}{x(x - 6)(x + 6)} = 0$
$\frac{-2(x - 6)}{x(x - 6)(x + 6)} = 0$
$\frac{-2}{x(x + 6)} = 0$
−2 ≠ 0
Ответ: нет корней