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

$\frac{2y - 2}{y + 3} + \frac{y + 3}{y - 3} = 5$ |*(y + 3)(y − 3)
$\begin{equation*} \begin{cases} (2y - 2)(y - 3) + (y + 3)^2 = 5(y + 3)(y - 3) &\\ x ≠ ±3 & \end{cases} \end{equation*}$
$2y^2 - 8y + 6 + y^2 + 6y + 9 = 5(y^2 - 9)$
$2y^2 + 2y - 60 = 0$
$y^2 + y - 30 = 0$
(y + 6)(y − 5) = 0
$y_1 = -6$
$y_2 = 5$
$\begin{equation*} \begin{cases} y_1 = -6, y_2 = 5 &\\ x ≠ ±3 & \end{cases} \end{equation*}$
Ответ:
$y_1 = -6$;
$y_2 = 5$.

$\frac{4}{9y^2 - 1} - \frac{4}{3y + 1} = \frac{5}{1 - 3y}$
$\frac{4}{9y^2 - 1} - \frac{4}{3y + 1} = -\frac{5}{3y - 1}$
$\frac{4}{9y^2 - 1} - \frac{4}{3y + 1} + \frac{5}{3y - 1} = 0$ |*(3y + 1)(3y − 1)
$\begin{equation*} \begin{cases} 4 - 4(3y - 1) + 5(3y + 1) = 0 &\\ y ≠ ±\frac{1}{3} & \end{cases} \end{equation*}$
4 − 12y + 4 + 15y + 5 = 0
3y + 13 = 0
$y = -4\frac{1}{3}$
$\begin{equation*} \begin{cases} y = -4\frac{1}{3} &\\ y ≠ ±\frac{1}{3} & \end{cases} \end{equation*}$
Ответ: $y = -4\frac{1}{3}$

$\frac{4}{x + 3} - \frac{5}{3 - x} = \frac{1}{x - 3} - 1$
$\frac{4}{x + 3} - \frac{5}{3 - x} - \frac{1}{x - 3} + 1 = 0$
$\frac{4}{x + 3} + \frac{4}{x - 3} + 1 = 0$ |*(x + 3)(x − 3)
$\begin{equation*} \begin{cases} 4(x - 3) + 4(x + 3) + x^2 - 9 = 0 &\\ x ≠ ±3 & \end{cases} \end{equation*}$
$x^2 + 8x - 9 = 0$
(x + 9)(x − 1) = 0
$x_1 = -9$
$x_2 = 1$
$\begin{equation*} \begin{cases} x_1 = -9, x_2 = 1 &\\ x ≠ ±3 & \end{cases} \end{equation*}$
Ответ:
$x_1 = -9$;
$x_2 = 1$.

$\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x^2 - x}$ |*x(x − 1)
$\begin{equation*} \begin{cases} 3(x - 1) + 4x = 5 - x &\\ x ≠ {0;1} & \end{cases} \end{equation*}$
3(x − 1) + 4x = 5 − x
3x − 3 + 4x + x = 5
8x = 5 + 3
8x = 8
x = 1
$\begin{equation*} \begin{cases} x = 1 &\\ x ≠ {0;1} & \end{cases} \end{equation*}$
x = ∅ − пустое множество, решений нет.

$\frac{3y - 2}{y} - \frac{1}{y - 2} = \frac{3y + 4}{y^2 - 2y}$ |*y(y − 2)
$\begin{equation*} \begin{cases} (3y - 2)(y - 2) - y = 3y + 4 &\\ y ≠ {0;2} & \end{cases} \end{equation*}$
$3^2 - 8y + 4 - y = 3y + 4$
$3y^2 - 12y = 0$
3y(y − 4) = 0
$y_1 = 0$
$y_2 = 4$
$\begin{equation*} \begin{cases} y_1 = 0, y_2 = 4 &\\ y ≠ {0;2} & \end{cases} \end{equation*}$
Ответ: y = 4