$\frac{y^2}{y + 3} = \frac{y}{y + 3}$
$\frac{y^2}{y + 3} - \frac{y}{y + 3} = 0$
$\frac{y^2 - y}{y + 3} = 0$
$\begin{equation*} \begin{cases} y^2 - y = 0 &\\ y + 3 ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y(y - 1) = 0 &\\ y ≠ -3 & \end{cases} \end{equation*}$
y = 0
или
y − 1 = 0
y = 1
Ответ: у = {0;1}

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

$\frac{2x^2}{x - 2} = \frac{-7x + 6}{2 - x}$
$\frac{2x^2}{x - 2} - \frac{-7x + 6}{2 - x} = 0$
$\frac{2x^2}{x - 2} + \frac{-7x + 6}{x - 2} = 0$
$\frac{2x^2 - 7x + 6}{x - 2} = 0$
$\begin{equation*} \begin{cases} 2x^2 - 7x + 6 = 0 &\\ x - 2 ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} (2x - 3)(x - 2) = 0 &\\ x - 2 ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x = {1,5;2} &\\ x ≠ 2 & \end{cases} \end{equation*}$
Ответ: x = 1,5

$\frac{y^2 - 6y}{y - 5} = \frac{5}{5 - y}$
$\frac{y^2 - 6y}{y - 5} - \frac{5}{5 - y} = 0$
$\frac{y^2 - 6y}{y - 5} + \frac{5}{y - 5} = 0$
$\frac{y^2 - 6y + 5}{y - 5} = 0$
$\begin{equation*} \begin{cases} y^2 - 6y + 5 = 0 &\\ y - 5 ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} (y - 1)(y - 5) = 0 &\\ y - 5 ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y = {1;5} &\\ y ≠ 5 & \end{cases} \end{equation*}$
Ответ: y = 1

$\frac{2x - 1}{x + 7} = \frac{3x + 4}{x - 1}$
$\frac{2x - 1}{x + 7} - \frac{3x + 4}{x - 1} = 0$
$\frac{(2x - 1)(x- 1) - (3x + 4)(x + 7)}{(x + 7)(x - 1)} = 0$
$\frac{2x^2 - 3x + 1 - (3x^2 + 25x + 28)}{(x + 7)(x - 1)} = 0$
$\frac{-x^2 - 28x - 27}{(x + 7)(x - 1)} = 0$
$\frac{-x^2 - 28x - 27}{(x + 7)(x - 1)}$
$\begin{equation*} \begin{cases} -x^2 - 28x - 27 = 0 &\\ (x + 7)(x - 1) ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x^2 + 28x + 27 = 0 &\\ x ≠ {-7;1} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} (x + 27)(x + 1) = 0 &\\ x ≠ {-7;1} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x = {-27;-1} &\\ x ≠ {-7;11} & \end{cases} \end{equation*}$
Ответ: x = {−27;−1}

$\frac{2y + 3}{2y - 1} = \frac{y - 5}{y + 3}$
$\frac{2y + 3}{2y - 1} - \frac{y - 5}{y + 3} = 0$
$\frac{(2y + 3)(y + 3) - (y - 5)(2y - 1)}{(2y - 1)(y + 3)}$
$\frac{2y^2 + 9y + 9 - (2y^2 - 11y + 5)}{(2y - 1)(y + 3)} = 0$
$\frac{20y + 4}{(2y - 1)(y + 3)}$
$\begin{equation*} \begin{cases} 20y + 4 = 0 &\\ (2y - 1)(y + 3) ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 20y = -4 &\\ (2y - 1)(y + 3) ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y = -0,2 &\\ y ≠ {-3; 0,5} & \end{cases} \end{equation*}$
Ответ: y = −0,2

$\frac{5y + 1}{y + 1} = \frac{y + 2}{y}$
$\frac{5y + 1}{y + 1} - \frac{y + 2}{y} = 0$
$\frac{(5y + 1)y - (y + 2)(y + 1)}{(y + 1)y} = 0$
$\frac{5y^2 + y - (y^2 + 3y + 2)}{(y + 1)y} = 0$
$\frac{4y^2 - 2y - 2}{(y + 1)y} = 0$
$\begin{equation*} \begin{cases} 4y^2 - 2y - 2 = 0 &\\ (y + 1)y ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 2y^2 - y - 1 = 0 &\\ y ≠ {-1;0} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} (2y + 1)(y - 1) = 0 &\\ y ≠ {-1;0} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y = {-0,5;1} &\\ y ≠ {-1;0} & \end{cases} \end{equation*}$
Ответ: y = {−0,5;1}

$\frac{1 + 3x}{1 - 2x} = \frac{5 - 3x}{1 + 2x}$
$\frac{1 + 3x}{1 - 2x} - \frac{5 - 3x}{1 + 2x} = 0$
$\frac{(1 + 3x)(1 + 2x) - (5 - 3x)(1 - 2x)}{(1 - 2x)(1 + 2x)} = 0$
$\frac{1 + 5x + 6x^2 - (5 - 13x + 6x^2)}{(1 - 2x)(1 + 2x)} = 0$
$\frac{18x - 4}{(1 - 2x)(1 + 2x)} = 0$
$\begin{equation*} \begin{cases} 18x - 4 = 0 &\\ (1 - 2x)(1 + 2x) ≠ 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x = \frac{2}{9} &\\ x ≠ ±\frac{1}{2} & \end{cases} \end{equation*}$
Ответ: $x = \frac{2}{9}$

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