$\frac{x - 1}{x + 5} + \frac{x + 5}{x - 1} = \frac{10}{3}$
x + 5 ≠ 0
x ≠ −5
и
x − 1 ≠ 0
x ≠ 1
$\frac{x - 1}{x + 5} + \frac{x + 5}{x - 1} - \frac{10}{3} = 0$ | * 3(x + 5)(x − 1)
$3(x - 1)^2 + 3(x + 5)^2 - 10(x + 5)(x - 1) = 0$
$3(x^2 - 2x + 1) + 3(x^2 + 10x + 25) - 10(x^2 + 5x - x - 5) = 0$
$3x^2 - 6x + 3 + 3x^2 + 30x + 75 - 10x^2 - 50x + 10x + 50 = 0$
$-4x^2 - 16x + 128 = 0$ | : (−4)
$x^2 + 4x - 32 = 0$
$D = b^2 - 4ac = 4^2 - 4 * 1 * (-32) = 16 + 128 = 144 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + \sqrt{144}}{2 * 1} = \frac{-4 + 12}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - \sqrt{144}}{2 * 1} = \frac{-4 - 12}{2} = \frac{-16}{2} = -8$
Ответ: −8 и 4

$\frac{x^2 - 3x + 6}{x} + \frac{2x}{x^2 - 3x + 6} = 3$
x ≠ 0
и
$x^2 - 3x + 6 ≠ 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 6 = 9 - 24 = -15 < 0$ − нет корней.
$y = \frac{x^2 - 3x + 6}{x} ≠ 0$
$y + \frac{2}{y} = 3$ | * y
$y^2 + 2 = 3y$
$y^2 - 3y + 2 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 1} = \frac{3 + 1}{2} = \frac{4}{2} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 1} = \frac{3 - 1}{2} = \frac{2}{2} = 1$
$\frac{x^2 - 3x + 6}{x} = 1$
или
$\frac{x^2 - 3x + 6}{x} = 2$
а)
$\frac{x^2 - 3x + 6}{x} = 1$ | * x
$x^2 - 3x + 6 = x$
$x^2 - 3x + 6 - x = 0$
$x^2 - 4x + 6 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 6 = 16 - 24 = -8 < 0$ − нет корней
б)
$\frac{x^2 - 3x + 6}{x} = 2$ | * x
$x^2 - 3x + 6 = 2x$
$x^2 - 3x + 6 - 2x = 0$
$x^2 - 5x + 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * 6 = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{1}}{2 * 1} = \frac{5 + 1}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{1}}{2 * 1} = \frac{5 - 1}{2} = \frac{4}{2} = 2$
Ответ: 2 и 3

$\frac{x^2}{(3x - 1)^2} - \frac{4x}{3x - 1} - 5 = 0$
3x − 1 ≠ 0
3x ≠ 1
$x ≠ \frac{1}{3}$
$y = \frac{x}{3x - 1}$
$y^2 - 4y - 5 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * (-5) = 16 + 20 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{36}}{2 * 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{36}}{2 * 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$
$\frac{x}{3x - 1} = 5$
или
$\frac{x}{3x - 1} = -1$
а)
$\frac{x}{3x - 1} = 5$
x = 5(3x − 1)
x = 15x − 5
x − 15x = −5
−14x = −5
$x = \frac{5}{14}$
б)
$\frac{x}{3x - 1} = -1$
x = −(3x − 1)
x = −3x + 1
x + 3x = 1
4x = 1
$x = \frac{1}{4}$
Ответ: $x = \frac{1}{4}$ и $x = \frac{5}{14}$

$\frac{24}{x^2 + 2x - 8} - \frac{15}{x^2 + 2x - 3} = 2$
$x^2 + 2x - 8 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * 8 = 4 + 32 = 36 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{36}}{2 * 1} = \frac{-2 + 6}{2} = \frac{4}{2} ≠ 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{36}}{2 * 1} = \frac{-2 - 6}{2} = \frac{-8}{2} ≠ -4$
и
$x^2 + 2x - 3 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} ≠ 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} ≠ -3$
$y = x^2 + 2x$
$\frac{24}{y - 8} - \frac{15}{y - 3} = 2$ | * (y − 8)(y − 3)
24(y − 3) − 15(y − 8) = 2(y − 8)(y − 3)
$24y - 72 - 15y + 120 = 2(y^2 - 8y - 3y + 24)$
$9y + 48 = 2y^2 - 16y - 6y + 48$
$9y + 48 = 2y^2 - 22y + 48$
$-2y^2 + 9y + 22y + 48 - 48 = 0$
$-2y^2 + 31y = 0$
−y(2y − 31) = 0
−y = 0
y = 0
или
2y − 31 = 0
2y = 31
y = 15,5
$x^2 + 2x = 0$
или
$x^2 + 2x = 15,5$
а)
$x^2 + 2x = 0$
x(x + 2) = 0
x = 0
или
x + 2 = 0
x = −2
б)
$x^2 + 2x = 15,5$
$x^2 + 2x - 15,5 = 0$ | * 2
$2x^2 + 4x - 31 = 0$
$D = b^2 - 4ac = 4^2 - 4 * 2 * (-31) = 16 + 248 = 264 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + \sqrt{264}}{2 * 2} = \frac{-4 + \sqrt{4 * 66}}{4} = \frac{-4 + 2\sqrt{66}}{4} = \frac{2(-2 + \sqrt{66})}{4} = \frac{-2 + \sqrt{66}}{2} = -1 + \frac{\sqrt{66}}{2}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - \sqrt{264}}{2 * 2} = \frac{-4 - \sqrt{4 * 66}}{4} = \frac{-4 - 2\sqrt{66}}{4} = \frac{2(-2 - \sqrt{66})}{4} = \frac{-2 - \sqrt{66}}{2} = -1 - \frac{\sqrt{66}}{2}$
Ответ: $-2; 0; -1 - \frac{\sqrt{66}}{2}; -1 + \frac{\sqrt{66}}{2}$.