$\frac{21}{x + 1} = \frac{16}{x - 2} - \frac{6}{x}$ |*x(x − 2)(x + 1)
$\begin{equation*} \begin{cases} 21x(x - 2) = 16x(x + 1) - 6(x - 2)(x + 1) &\\ x ≠ {-1;0;2} & \end{cases} \end{equation*}$
$21x^2 - 42x = 16x^2 + 16x - 6(x^2 - x - 2)$
$11x^2 - 64x - 12 = 0$
(11x + 2)(x − 6) = 0
$x_1 = -\frac{2}{11}$
$x_2 = 6$
$\begin{equation*} \begin{cases} x_1 = -\frac{2}{11}, x_2 = 6 &\\ x ≠ {-1;0;2} & \end{cases} \end{equation*}$
Ответ:
$x_1 = -\frac{2}{11}$;
$x_2 = 6$.

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

$\frac{18}{4x^2 + 4x + 1} - \frac{1}{2x^2 - x} = \frac{6}{4x^2 - 1}$
$\frac{18}{(2x + 1)^2} - \frac{1}{x(2x - 1)} = \frac{6}{(2x - 1)(2x + 1)} |* x(2x + 1)^2(2x - 1)$
$\begin{equation*} \begin{cases} 18x(2x - 1) - (2x + 1)^2 = 6x(2x + 1) &\\ y ≠ {0;±\frac{1}{2}} & \end{cases} \end{equation*}$
$36x^2 - 18x - 4x^2 - 4x - 1 = 12x^2 + 6x$
$20x^2 - 28x - 1 = 0$
$D = 14^2 + 20 = 216 = (6\sqrt{6})^2$
$x = \frac{14 ± 6\sqrt{54}}{20} = \frac{7 ± 3\sqrt{6}}{10}$
$\begin{equation*} \begin{cases} x = \frac{7 ± 3\sqrt{6}}{10} &\\ x ≠ {0;±\frac{1}{2}} & \end{cases} \end{equation*}$
Ответ: $x = \frac{7 ± 3\sqrt{6}}{10}$

$\frac{3(4y^2 + 10y - 7)}{16y^2 - 9} = \frac{3y - 7}{3 - 4y} + \frac{6y + 5}{3 + 4y}$
$\frac{3(4y^2 + 10y - 7)}{16y^2 - 9} = -\frac{3y - 7}{4y - 3} + \frac{6y + 5}{4y + 3}$ |* (4y − 3)(4y + 3)
$\begin{equation*} \begin{cases} 3(4y^2 + 10y - 7) = -(3y - 7)(4y + 3) + (6y + 5)(4y - 3) &\\ y ≠ ±\frac{3}{4} & \end{cases} \end{equation*}$
$12y^2 + 30y - 21 = -(12y^2 - 19y - 21) + 24y^2 + 2y - 15$
$12y^2 + 30y - 21 = -12y^2 + 19y + 21 + 24y^2 + 2y - 15$
9y = 27
y = 3
$\begin{equation*} \begin{cases} y = 3 &\\ y ≠ ±\frac{3}{4} & \end{cases} \end{equation*}$
Ответ: y = 3