$\frac{10}{x + 2} + \frac{9}{x} = 1$
x + 2 ≠ 0
x ≠ −2
и
x ≠ 0
$\frac{10}{x + 2} + \frac{9}{x} - 1 = 0$ | * x(x + 2)
10x + 9(x + 2) − x(x + 2) = 0
$10x + 9x + 18 - x^2 - 2x = 0$
$-x^2 + 17x + 18 = 0$ | * 1
$x^2 - 17x - 18 = 0$
$D = b^2 - 4ac = (-17)^2 - 4 * 1 * (-18) = 289 + 72 = 361 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{17 + \sqrt{361}}{2 * 1} = \frac{17 + 19}{2} = \frac{36}{2} = 18$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{17 - \sqrt{361}}{2 * 1} = \frac{17 - 19}{2} = \frac{-2}{2} = -1$
Ответ: −1 и 18

$\frac{48}{14 - x} - \frac{48}{14 + x} = 1$
14 − x ≠ 0
x ≠ 14
и
14 + x ≠ 0
x ≠ −14
$\frac{48}{14 - x} - \frac{48}{14 + x} - 1 = 0$ | * (14 − x)(14 + x)
48(14 + x) − 48(14 − x) − (14 − x)(14 + x) = 0
$672 + 48x - 672 + 48x - (196 - x^2) = 0$
$x^2 + 96x - 196 = 0$
$D = b^2 - 4ac = 96^2 - 4 * 1 * (-196) = 9216 + 784 = 10000 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-96 + \sqrt{10000}}{2 * 1} = \frac{-96 + 100}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-96 - \sqrt{10000}}{2 * 1} = \frac{-96 - 100}{2} = \frac{-196}{2} = -98$
Ответ: −98 и 2

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

$\frac{x - 1}{x + 3} + \frac{x + 1}{x - 3} = \frac{2x + 18}{x^2 - 9}$
x + 3 ≠ 0
x ≠ −3
и
x − 3 ≠ 0
x ≠ 3
$\frac{x - 1}{x + 3} + \frac{x + 1}{x - 3} - \frac{2x + 18}{(x - 3)(x + 3)} = 0$ | * (x − 3)(x + 3)
(x − 1)(x − 3) + (x + 1)(x + 3) − (2x + 18) = 0
$x^2 - x - 3x + 3 + x^2 + x + 3x + 3 - 2x - 18 = 0$
$2x^2 - 2x - 12 = 0$ | : 2
$x^2 - x - 6 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{25}}{2 * 1} = \frac{1 + 5}{2} = \frac{6}{2} = 3$ − не удовлетворяет условию, так как x ≠ 3.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{25}}{2 * 1} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$
Ответ: −2

$\frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} = 4$
x − 1 ≠ 0
x ≠ 1
и
x + 1 ≠ 0
x ≠ −1
$\frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} - 4 = 0$ | * (x − 1)(x + 1)
$(4x - 10)(x + 1) + (x + 6)(x - 1) - 4(x^2 - 1) = 0$
$4x^2 - 10x + 4x - 10 + x^2 + 6x - x - 6 - 4x^2 + 4 = 0$
$x^2 - x - 12 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-12) = 1 + 48 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{49}}{2 * 1} = \frac{1 + 7}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{49}}{2 * 1} = \frac{1 - 7}{2} = \frac{-6}{2} = -3$
Ответ: −3 и 4

$\frac{1}{x} - \frac{10}{x^2 - 5x} = \frac{3 - x}{x - 5}$
$\frac{1}{x} - \frac{10}{x(x - 5)} = \frac{3 - x}{x - 5}$
x ≠ 0
и
x − 5 ≠ 0
x ≠ 5
$\frac{1}{x} - \frac{10}{x(x - 5)} - \frac{3 - x}{x - 5} = 0$ | * x(x − 5)
x − 5 − 10 − x(3 − x) = 0
$x - 15 - 3x + x^2 = 0$
$x^2 - 2x - 15 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-15) = 4 + 60 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{64}}{2 * 1} = \frac{2 + 8}{2} = \frac{10}{2} = 5$ − не удовлетворяет условию, так как x ≠ 5.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{64}}{2 * 1} = \frac{2 - 8}{2} = \frac{-6}{2} = -3$
Ответ: −3

$\frac{4x}{x^2 + 4x + 4} - \frac{x - 2}{x^2 + 2x} = \frac{1}{x}$
$\frac{4x}{(x + 2)^2} - \frac{x - 2}{x(x + 2)} = \frac{1}{x}$
x ≠ 0
и
$(x + 2)^2 ≠ 0$
x + 2 ≠ 0
x ≠ −2
$\frac{4x}{(x + 2)^2} - \frac{x - 2}{x(x + 2)} - \frac{1}{x} = 0$ | * $x(x + 2)^2$
$4x * x - (x - 2)(x + 2) - (x + 2)^2 = 0$
$4x^2 - (x^2 - 4) - (x^2 + 4x + 4) = 0$
$4x^2 - x^2 + 4 - x^2 - 4x - 4 = 0$
$2x^2 - 4x = 0$
2x(x − 2) = 0
2x = 0
x = 0 − не удовлетворяет условию, так как x ≠ 0.
или
x − 2 = 0
x = 2
Ответ: 2

$\frac{6}{x^2 - 36} - \frac{3}{x^2 - 6x} + \frac{x - 12}{x^2 + 6x} = 0$
$\frac{6}{(x - 6)(x + 6)} - \frac{3}{x(x - 6)} + \frac{x - 12}{x(x + 6)} = 0$
x ≠ 0
и
x − 6 ≠ 0
x ≠ 6
и
x + 6 ≠ 0
x ≠ −6
$\frac{6}{(x - 6)(x + 6)} - \frac{3}{x(x - 6)} + \frac{x - 12}{x(x + 6)} = 0$ | * x(x − 6)(x + 6)
$6x - 3(x + 6) + (x - 12)(x - 6) = 0$
$6x - 3x - 18 + x^2 - 12x - 6x + 72 = 0$
$x^2 - 15x + 54 = 0$
$D = b^2 - 4ac = (-15)^2 - 4 * 1 * 54 = 225 - 216 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{15 + \sqrt{9}}{2 * 1} = \frac{15 + 3}{2} = \frac{18}{2} = 9$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{15 - \sqrt{9}}{2 * 1} = \frac{15 - 3}{2} = \frac{12}{2} = 6$ − не удовлетворяет условию, так как x ≠ 6.
Ответ: 9

$\frac{x}{x + 7} + \frac{x + 7}{x - 7} = \frac{63 - 5x}{x^2 - 49}$
$\frac{x}{x + 7} + \frac{x + 7}{x - 7} - \frac{63 - 5x}{(x - 7)(x + 7)} = 0$
x − 7 ≠ 0
x ≠ 7
и
x + 7 ≠ 0
x ≠ −7
$\frac{x}{x + 7} + \frac{x + 7}{x - 7} - \frac{63 - 5x}{(x - 7)(x + 7)} = 0$ | * (x − 7)(x + 7)
$x(x - 7) + (x + 7)^2 - (63 - 5x) = 0$
$x^2 - 7x + x^2 + 14x + 49 - 63 + 5x = 0$
$2x^2 + 12x - 14 = 0$ | : 2
$x^2 + 6x - 7 = 0$
$D = b^2 - 4ac = 6^2 - 4 * 1 * (-7) = 36 + 28 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{64}}{2 * 1} = \frac{-6 + 8}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{64}}{2 * 1} = \frac{-6 - 8}{2} = \frac{-14}{2} = -7$ − не удовлетворяет условию, так как x ≠ −7.
Ответ: 1

$\frac{4}{x^2 - 10x + 25} - \frac{1}{x + 5} = \frac{10}{x^2 - 25}$
$\frac{4}{(x - 5)^2} - \frac{1}{x + 5} = \frac{10}{(x - 5)(x + 5)}$
$(x - 5)^2 ≠ 0$
x − 5 ≠ 0
x ≠ 5
и
x + 5 ≠ 0
x ≠ −5
$\frac{4}{(x - 5)^2} - \frac{1}{x + 5} - \frac{10}{(x - 5)(x + 5)} = 0$ | * $(x - 5)^2(x + 5)$
$4(x + 5) - (x - 5)^2 - 10(x - 5) = 0$
$4x + 20 - (x^2 - 10x + 25) - 10x + 50 = 0$
$4x + 20 - x^2 + 10x - 25 - 10x + 50 = 0$
$-x^2 + 4x + 45 = 0$ | * (−1)
$x^2 - 4x - 45 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * (-45) = 16 + 180 = 196 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{196}}{2 * 1} = \frac{4 + 14}{2} = \frac{18}{2} = 9$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{196}}{2 * 1} = \frac{4 - 14}{2} = \frac{-10}{2} = -5$ − не удовлетворяет условию, так как x ≠ −5.
Ответ: 9