$\frac{2x - 10}{x^3 + 1} + \frac{4}{x + 1} = \frac{5x - 1}{x^2 - x + 1}$
$\frac{2x - 10}{(x + 1)(x^2 - x + 1)} + \frac{4}{x + 1} = \frac{5x - 1}{x^2 - x + 1}$
x + 1 ≠ 0
x ≠ −1
и
$x^2 - x + 1 ≠ 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * 1 = 1 - 4 = -3 < 0$ − нет корней.
$\frac{2x - 10}{(x + 1)(x^2 - x + 1)} + \frac{4}{x + 1} - \frac{5x - 1}{x^2 - x + 1} = 0$ | * $(x + 1)(x^2 - x + 1)$
$2x - 10 + 4(x^2 - x + 1) - (5x - 1)(x + 1) = 0$
$2x - 10 + 4x^2 - 4x + 4 - (5x^2 - x + 5x - 1) = 0$
$4x^2 - 2x - 6 - 5x^2 - 4x + 1 = 0$
$-x^2 - 6x - 5 = 0$ | * (−1)
$x^2 + 6x + 5 = 0$
$D = b^2 - 4ac = 6^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{16}}{2 * 1} = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$ − не удовлетворяет условию, так как x ≠ −1.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{16}}{2 * 1} = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$
Ответ: −5

$\frac{6}{x^2 - 4x + 3} + \frac{5 - 2x}{x - 1} = \frac{3}{x - 3}$
x − 1 ≠ 0
x ≠ 1
и
x − 3 ≠ 0
x ≠ 3
и
$x^2 - 4x + 3 ≠ 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 3 = 16 - 12 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{4}}{2 * 1} = \frac{4 + 2}{2} = \frac{6}{2} ≠ 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{4}}{2 * 1} = \frac{4 - 2}{2} = \frac{2}{2} ≠ 1$
$x^2 - 4x + 3 = (x - 3)(x - 1)$
$\frac{6}{(x - 3)(x - 1)} + \frac{5 - 2x}{x - 1} - \frac{3}{x - 3} = 0$ | * (x − 3)(x − 1)
$6 + (5 - 2x)(x - 3) - 3(x - 1) = 0$
$6 + 5x - 2x^2 - 15 + 6x - 3x + 3 = 0$
$-2x^2 + 8x - 6 = 0$ | : (−2)
$x^2 - 4x + 3 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 3 = 16 - 12 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{4}}{2 * 1} = \frac{4 + 2}{2} = \frac{6}{2} = 3$ − не удовлетворяет условию, так как x ≠ 3.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{4}}{2 * 1} = \frac{4 - 2}{2} = \frac{2}{2} = 1$ − не удовлетворяет условию, так как x ≠ 1.
Ответ: нет корней

$\frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{14}{x^2 + 3x + 2}$
x + 2 ≠ 0
x ≠ −2
и
x + 1 ≠ 0
x ≠ −1
и
$x^2 + 3x + 2 ≠ 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{1}}{2 * 1} = \frac{-3 + 1}{2} = \frac{-2}{2} ≠ -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{1}}{2 * 1} = \frac{-3 - 1}{2} = \frac{-4}{2} ≠ -2$
$x^2 + 3x + 2 = (x - (-1))(x - (-2)) = (x + 1)(x + 2)$
$\frac{4x - 6}{x + 2} - \frac{x}{x + 1} - \frac{14}{(x + 1)(x + 2)} = 0$ | * (x + 1)(x + 2)
$(4x - 6)(x + 1) - x(x + 2) - 14 = 0$
$4x^2 - 6x + 4x - 6 - x^2 - 2x - 14 = 0$
$3x^2 - 4x - 20 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 3 * (-20) = 16 + 240 = 256 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{256}}{2 * 3} = \frac{4 + 16}{6} = \frac{20}{6} = \frac{10}{3} = 3\frac{1}{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{256}}{2 * 3} = \frac{4 - 16}{6} = \frac{-12}{6} = -4$
Ответ: −4 и $3\frac{1}{3}$

$\frac{x}{x^2 - 4} - \frac{3x - 1}{x^2 + x - 6} = \frac{2}{x^2 + 5x + 6}$
$x^2 - 4 ≠ 0$
(x − 2)(x + 2) ≠ 0
x − 2 ≠ 0
x ≠ 2
и
x + 2 ≠ 0
x ≠ −2
и
$x^2 + x - 6 ≠ 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 * 1} = \frac{-1 + 5}{2} = \frac{4}{2} ≠ 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 * 1} = \frac{-1 - 5}{2} = \frac{-6}{2} ≠ -3$
$x^2 + x - 6 = (x - 2)(x - (-3)) = (x - 2)(x + 3)$
$x^2 + 5x + 6 ≠ 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 6 = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{1}}{2 * 1} = \frac{-5 + 1}{2} = \frac{-4}{2} ≠ -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{1}}{2 * 1} = \frac{-5 - 1}{2} = \frac{-6}{2} ≠ -3$
$x^2 + 5x + 6 = (x - (-2))(x - (-3)) = (x + 2)(x + 3)$
$\frac{x}{(x - 2)(x + 2)} - \frac{3x - 1}{(x - 2)(x + 3)} - \frac{2}{(x + 2)(x + 3)} = 0$ | * (x − 2)(x + 2)(x + 3)
x(x + 3) − (3x − 1)(x + 2) − 2(x − 2) = 0
$x^2 + 3x - (3x^2 - x + 6x - 2) - 2x + 4 = 0$
$x^2 + 3x - 3x^2 + x - 6x + 2 - 2x + 4 = 0$
$-2x^2 - 4x + 6 = 0$ | : (−2)
$x^2 + 2x - 3 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$ − не удовлетворяет условию, так как x ≠ −3.
Ответ: 1