$\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{x^3 - 8} = \frac{5}{x - 2}$
$\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{(x - 2)(x^2 + 2x + 4)} = \frac{5}{x - 2}$
x − 2 ≠ 0
x ≠ 2
и
$x^2 + 2x + 4 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * 4 = 4 - 16 = -12 < 0$ − нет корней
$\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{(x - 2)(x^2 + 2x + 4)} - \frac{5}{x - 2} = 0$ | * $(x - 2)(x^2 + 2x + 4)$
$(3x + 2)(x - 2) + x^2 + 39 - 5(x^2 + 2x + 4) = 0$
$3x^2 + 2x - 6x - 4 + x^2 + 39 - 5x^2 - 10x - 20 = 0$
$-x^2 - 14x + 15 = 0$ | * (−1)
$x^2 + 14x - 15 = 0$
$D = b^2 - 4ac = 14^2 - 4 * 1 * (-15) = 196 + 60 = 256 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-14 + \sqrt{256}}{2 * 1} = \frac{-14 + 16}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-14 - \sqrt{256}}{2 * 1} = \frac{-14 - 16}{2} = \frac{-30}{2} = -15$
Ответ: −15 и 1

$\frac{x}{x - 1} + \frac{x + 1}{x + 3} = \frac{8}{x^2 + 2x - 3}$
x − 1 ≠ 0
x ≠ 1
и
x + 3 ≠ 0
x ≠ −3
и
$x^2 + 2x - 3 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} ≠ 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} ≠ -3$
$x^2 + 2x - 3 = (x - 1)(x - (-3)) = (x - 1)(x + 3)$
$\frac{x}{x - 1} + \frac{x + 1}{x + 3} - \frac{8}{(x - 1)(x + 3)} = 0$ | * (x − 1)(x + 3)
x(x + 3) + (x + 1)(x − 1) − 8 = 0
$x^2 + 3x + x^2 + x - x - 1 - 8 = 0$
$2x^2 + 3x - 9 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 2 * (-9) = 9 + 72 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{81}}{2 * 2} = \frac{-3 + 9}{4} = \frac{6}{4} = 1,5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{81}}{2 * 2} = \frac{-3 - 9}{4} = \frac{-12}{4} = -3$ − не удовлетворяет условию, так как x ≠ −3.
Ответ: 1,5