$(x^2 - 6x)^2 + (x^2 - 6x) - 56 = 0$
$y = x^2 - 6x$
$y^2 + y - 56 = 0$
$D = b^2 - 4ac =1^2 - 4 * 1 * (-56) = 1 + 224 = 225 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{225}}{2 * 1} = \frac{-1 + 15}{2} = \frac{14}{2} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{225}}{2 * 1} = \frac{-1 - 15}{2} = \frac{-16}{2} = -8$
$x^2 - 6x = 7$
$x^2 - 6x - 7 = 0$
$D = b^2 - 4ac =(-6)^2 - 4 * 1 * (-7) = 36 + 28 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{64}}{2 * 1} = \frac{6 + 8}{2} = \frac{14}{2} = 7$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{64}}{2 * 1} = \frac{6 - 8}{2} = \frac{-2}{2} = -1$
или
$x^2 - 6x = -8$
$x^2 - 6x + 8 = 0$
$D = b^2 - 4ac =(-6)^2 - 4 * 1 * 8 = 36 - 32 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{4}}{2 * 1} = \frac{6 + 2}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{4}}{2 * 1} = \frac{6 - 2}{2} = \frac{4}{2} = 2$
Ответ: −1, 2, 4, 7.

$(x^2 + 8x + 3)(x^2 + 8x + 5) = 63$
$y = x^2 + 8x$
$(y + 3)(y + 5) = 63$
$y^2 + 3y + 5y + 15 - 63 = 0$
$y^2 + 8y - 48 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * (-48) = 64 + 192 = 256 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{256}}{2 * 1} = \frac{-8 + 16}{2} = \frac{8}{2} = 4$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{256}}{2 * 1} = \frac{-8 - 16}{2} = \frac{-24}{2} = -12$
$x^2 + 8x = 4$
$x^2 + 8x - 4 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * (-4) = 64 + 16 = 80 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{80}}{2 * 1} = \frac{-8 + \sqrt{16 * 5}}{2} = \frac{-8 + 4\sqrt{5}}{2} = -4 + 2\sqrt{5}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{80}}{2 * 1} = \frac{-8 - \sqrt{16 * 5}}{2} = \frac{-8 - 4\sqrt{5}}{2} = -4 - 2\sqrt{5}$
или
$x^2 + 8x = -12$
$x^2 + 8x + 12 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * 12 = 64 - 48 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{16}}{2 * 1} = \frac{-8 + 4}{2} = \frac{-4}{2} = -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{16}}{2 * 1} = \frac{-8 - 4}{2} = \frac{-12}{2} = -6$
Ответ: $-4 - 2\sqrt{5}, -4 + 2\sqrt{5}, -6, -2.$

$\frac{x^4}{(x - 2)^2} - \frac{4x^2}{x - 2} - 5 = 0$
$(x - 2)^2 ≠ 0$
x − 2 ≠ 0
x ≠ 2
$(\frac{x^2}{x - 2})^2 - \frac{4x^2}{x - 2} - 5 = 0$
$y = \frac{x^2}{x - 2}$
$y^2 - 4y - 5 = 0$
$D = b^2 - 4ac =(-4)^2 - 4 * 1 * (-5) = 16 + 20 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{36}}{2 * 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{36}}{2 * 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$
$\frac{x^2}{x - 2} = 5$ | * (x − 2)
$x^2 = 5(x - 2)$
$x^2 = 5x - 10$
$x^2 - 5x + 10 = 0$
$D = b^2 - 4ac =(-5)^2 - 4 * 1 * 10 = 25 - 40 = 15 < 0$ − нет корней
или
$\frac{x^2}{x - 2} = -1$ | * (x − 2)
$x^2 = -(x - 2)$
$x^2 = -x + 2$
$x^2 + x - 2 = 0$
$D = b^2 - 4ac =1^2 - 4 * 1 * (-2) = 1 + 8 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{9}}{2 * 1} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{9}}{2 * 1} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2$
Ответ: −2 и 1

$\frac{x + 4}{x - 3} - \frac{x - 3}{x + 4} = \frac{3}{2}$
x − 3 ≠ 0
x ≠ 3
и
x + 4 ≠ 0
x ≠ −4
$y = \frac{x + 4}{x - 3}$
$y - \frac{1}{y} - \frac{3}{2} = 0$ | * 2y
$2y^2 - 3y - 2 = 0$
$D = b^2 - 4ac =(-3)^2 - 4 * 2 * (-2) = 9 + 16 = 25 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{25}}{2 * 2} = \frac{3 + 5}{4} = \frac{8}{4} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{25}}{2 * 2} = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}$
$\frac{x + 4}{x - 3} = 2$
x + 4 = 2(x − 3)
x + 4 = 2x − 6
x − 2x = −6 − 4
−x = −10
x = 10
или
$\frac{x + 4}{x - 3} = -\frac{1}{2}$ | * 2(x − 3)
2(x + 4) = −(x − 3)
2x + 8 = −x + 3
2x + x = 3 − 8
3x = −5
$x = -\frac{5}{3} = -1\frac{2}{3}$
Ответ: $-1\frac{2}{3}$ и 10