$\frac{x^2 - 5x - 6}{x - 6} = 0$
x − 6 ≠ 0
x ≠ 6
$x^2 - 5x - 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * (-6) = 25 + 24 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{49}}{2 * 1} = \frac{5 + 7}{2} = \frac{12}{2} = 6$ − не удовлетворяет условию, так как x ≠ 6.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{49}}{2 * 1} = \frac{5 - 7}{2} = \frac{-2}{2} = -1$
Ответ: −1

$\frac{4x^2 - 7x - 2}{x - 2} = 0$
x − 2 ≠ 0
x ≠ 2
$D = b^2 - 4ac = (-7)^2 - 4 * 4 * (-2) = 49 + 32 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{81}}{2 * 4} = \frac{7 + 9}{8} = \frac{16}{8} = 2$ − не удовлетворяет условию, так как x ≠ 2.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{81}}{2 * 4} = \frac{7 - 9}{8} = \frac{-2}{8} = -\frac{1}{4}$
Ответ: $-\frac{1}{4}$

$\frac{2x^2 + 6}{x + 8} = \frac{13x}{x + 8}$
x + 8 ≠ 0
x ≠ −8
$\frac{2x^2 + 6}{x + 8} = \frac{13x}{x + 8}$ | * (x + 8)
$2x^2 + 6 = 13x$
$2x^2 - 13x + 6 = 0$
$D = b^2 - 4ac = (-13)^2 - 4 * 2 * 6 = 169 - 48 = 121 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{13 + \sqrt{121}}{2 * 2} = \frac{13 + 11}{4} = \frac{24}{4} = 6$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{13 - \sqrt{121}}{2 * 2} = \frac{13 - 11}{4} = \frac{2}{4} = \frac{1}{2}$
Ответ: $\frac{1}{2}$ и 6

$\frac{x^2 + 4x}{x + 7} = \frac{5x + 56}{x + 7}$
x + 7 ≠ 0
x ≠ −7
$\frac{x^2 + 4x}{x + 7} = \frac{5x + 56}{x + 7}$ | * (x + 7)
$x^2 + 4x = 5x + 56$
$x^2 + 4x - 5x - 56 = 0$
$x^2 - x - 56 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-56) = 1 + 224 = 225 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{225}}{2 * 1} = \frac{1 + 15}{2} = \frac{16}{2} = 8$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{225}}{2 * 1} = \frac{1 - 15}{2} = \frac{-14}{2} = -7$ − не удовлетворяет условию, так как x ≠ −7.
Ответ: 8

$\frac{x^2 + 12x}{x + 4} - \frac{5x - 12}{x + 4} = 0$
x + 4 ≠ 0
x ≠ −4
$\frac{x^2 + 12x}{x + 4} - \frac{5x - 12}{x + 4} = 0$ | * (x + 4)
$x^2 + 12x - (5x - 12) = 0$
$x^2 + 12x - 5x + 12 = 0$
$x^2 + 7x + 12 = 0$
$D = b^2 - 4ac = 7^2 - 4 * 1 * 12 = 49 - 48 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-7 + \sqrt{1}}{2 * 1} = \frac{-7 + 1}{2} = \frac{-6}{2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-7 - \sqrt{1}}{2 * 1} = \frac{-7 - 1}{2} = \frac{-8}{2} = -4$ − не удовлетворяет условию, так как x ≠ −4.
Ответ: −3

$\frac{x^2 - 3x}{x + 6} = 6$
x + 6 ≠ 0
x ≠ −6
$\frac{x^2 - 3x}{x + 6} = 6$ | * (x + 6)
$x^2 - 3x = 6(x + 6)$
$x^2 - 3x = 6x + 36$
$x^2 - 3x - 6x - 36 = 0$
$x^2 - 9x - 36 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 1 * (-36) = 81 + 144 = 225 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{225}}{2 * 1} = \frac{9 + 15}{2} = \frac{24}{2} = 12$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{225}}{2 * 1} = \frac{9 - 15}{2} = \frac{-6}{2} = -3$
Ответ: −3 и 12

$\frac{2 - 33y}{y - 4} = 7y$
y − 4 ≠ 0
y ≠ 4
$\frac{2 - 33y}{y - 4} = 7y$ | * (y − 4)
$2 - 33y = 7y(y - 4)$
$2 - 33y = 7y^2 - 28y$
$-7y^2 + 28y - 33y + 2 = 0$
$-7y^2 - 5y + 2 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * (-7) * 2 = 25 + 56 = 81 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{81}}{2 * (-7)} = \frac{5 + 9}{-14} = \frac{14}{-14} = -1$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{81}}{2 * (-7)} = \frac{5 - 9}{-14} = \frac{-4}{-14} = \frac{2}{7}$
Ответ: −1 и $\frac{2}{7}$

$y - \frac{39}{y} = 10$
y ≠ 0
$y - \frac{39}{y} = 10$ | * y
$y^2 - 39 = 10y$
$y^2 - 10y - 39 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 1 * (-39) = 100 + 156 = 256 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{256}}{2 * 1} = \frac{10 + 16}{2} = \frac{26}{2} = 13$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{256}}{2 * 1} = \frac{10 - 16}{2} = \frac{-6}{2} = -3$
Ответ: −3 и 13