$\frac{x^2 - 7x}{x + 1} = \frac{8}{x + 1}$
x + 1 ≠ 0
x ≠ −1
$\frac{x^2 - 7x}{x + 1} - \frac{8}{x + 1} = 0$ | * (x + 1)
$x^2 - 7x - 8 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 1 * (-8) = 49 + 32 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{81}}{2 * 1} = \frac{7 + 9}{2} = \frac{16}{2} = 8$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{81}}{2 * 1} = \frac{7 - 9}{2} = \frac{-2}{2} = -1$ − не является решением, так как x ≠ −1
Ответ: 8

$\frac{3x^2 + 4x}{x^2 - 9} = \frac{3 - 4x}{x^2 - 9}$
$x^2 - 9 ≠ 0$
(x − 3)(x + 3) ≠ 0
x − 3 ≠ 0
x ≠ 3
и
x + 3 ≠ 0
x ≠ −3
$\frac{3x^2 + 4x}{x^2 - 9} - \frac{3 - 4x}{x^2 - 9} = 0$ | * $(x^2 - 9)$
$3x^2 + 4x - (3 - 4x) = 0$
$3x^2 + 4x - 3 + 4x = 0$
$3x^2 + 8x - 3 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 3 * (-3) = 64 + 36 = 100 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{100}}{2 * 3} = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{100}}{2 * 3} = \frac{-8 - 10}{6} = \frac{-18}{6} = -3$ − не является решением, так как x ≠ −3
Ответ: $\frac{1}{3}$

$\frac{4 - x}{4x - 3} = \frac{2x - 2}{7 - x}$
4x − 3 ≠ 0
4x ≠ 3
$x ≠ \frac{3}{4}$
и
7 − x ≠ 0
x ≠ 7
$\frac{4 - x}{4x - 3} - \frac{2x - 2}{7 - x} = 0$ | * (4x − 3)(7 − x)
(4 − x)(7 − x) − (2x − 2)(4x − 3) = 0
$28 - 7x - 4x + x^2 - (8x^2 - 8x - 6x + 6) = 0$
$x^2 - 11x + 28 - 8x^2 + 8x + 6x - 6 = 0$
$-7x^2 + 3x + 22 = 0$ | * (−1)
$7x^2 - 3x - 22 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 7 * (-22) = 9 + 616 = 625 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{625}}{2 * 7} = \frac{3 + 25}{14} = \frac{28}{14} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{625}}{2 * 7} = \frac{3 - 25}{14} = \frac{-22}{14} = -\frac{11}{7} = -1\frac{4}{7}$
Ответ: $-1\frac{4}{7}$ и 2

$\frac{1}{x + 1} - \frac{1}{x - 6} = \frac{7}{12}$
x + 1 ≠ 0
x ≠ −1
и
x − 6 ≠ 0
x ≠ 6
$\frac{1}{x + 1} - \frac{1}{x - 6} - \frac{7}{12} = 0$ | * 12(x + 1)(x − 6)
12(x − 6) − 12(x + 1) − 7(x + 1)(x − 6) = 0
$12x - 72 - 12x - 12 - 7(x^2 + x - 6x - 6) = 0$
$-84 - 7x^2 - 7x + 42x + 42 = 0$
$-7x^2 + 35x - 42 = 0$ | : (−7)
$x^2 - 5x + 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * 6 = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{1}}{2 * 1} = \frac{5 + 1}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{1}}{2 * 1} = \frac{5 - 1}{2} = \frac{4}{2} = 2$
Ответ: 2 и 3

$\frac{63}{x^2 + 3x} - \frac{2}{x^2 - 3x} = \frac{7}{x}$
$x^2 + 3x ≠ 0$
x(x + 3) ≠ 0
x ≠ 0
и
x + 3 ≠ 0
x ≠ −3
и
$x^2 - 3x ≠ 0$
x(x − 3) ≠ 0
x ≠ 0
и
x − 3 ≠ 0
x ≠ 3
$\frac{63}{x(x + 3)} - \frac{2}{x(x - 3)} = \frac{7}{x}$
$\frac{63}{x(x + 3)} - \frac{2}{x(x - 3)} - \frac{7}{x} = 0$ | * x(x + 3)(x − 3)
63(x − 3) − 2(x + 3) − 7(x + 3)(x − 3) = 0
$63x - 189 - 2x - 6 - 7(x^2 - 9) = 0$
$61x - 195 - 7x^2 + 63 = 0$
$-7x^2 + 61x - 132 = 0$ | * (−1)
$7x^2 - 61x + 132 = 0$
$D = b^2 - 4ac = (-61)^2 - 4 * 7 * 132 = 3721 - 3696 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{61 + \sqrt{25}}{2 * 7} = \frac{61 + 5}{14} = \frac{66}{14} = \frac{33}{7} = 4\frac{5}{7}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{61 - \sqrt{25}}{2 * 7} = \frac{61 - 5}{14} = \frac{56}{14} = 4$
Ответ: 4 и $4\frac{5}{7}$

$\frac{2x}{x - 2} + \frac{3}{x + 4} = \frac{4x - 2}{(x + 4)(x - 2)}$
x − 2 ≠ 0
x ≠ 2
и
x + 4 ≠ 0
x ≠ −4
$\frac{2x}{x - 2} + \frac{3}{x + 4} - \frac{4x - 2}{(x + 4)(x - 2)} = 0$ | * (x + 4)(x − 2)
$2x(x + 4) + 3(x - 2) - (4x - 2) = 0$
$2x^2 + 8x + 3x - 6 - 4x + 2 = 0$
$2x^2 + 7x - 4 = 0$
$D = b^2 - 4ac = 7^2 - 4 * 2 * (-4) = 49 + 32 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-7 + \sqrt{81}}{2 * 2} = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-7 - \sqrt{81}}{2 * 2} = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$ − не является решением, так как x ≠ −4.
Ответ: $\frac{1}{2}$

$\frac{1}{x^2 + 2x} - \frac{2}{x^2 - 4} = \frac{x + 4}{5x(2 - x)}$
$\frac{1}{x(x + 2)} - \frac{2}{(x - 2)(x + 2)} = -\frac{x + 4}{5x(x - 2)}$
5x ≠ 0
x ≠ 0
и
x + 2 ≠ 0
x ≠ −2
и
x − 2 ≠ 0
x ≠ 2
$\frac{1}{x(x + 2)} - \frac{2}{(x - 2)(x + 2)} + \frac{x + 4}{5x(x - 2)} = 0$ | * 5x(x + 2)(x − 2)
5(x − 2) − 10x + (x + 4)(x + 2) = 0
$5x -10 - 10x + x^2 + 4x + 2x + 8 = 0$
$-5x - 10 + x^2 + 6x + 8 = 0$
$x^2 + x - 2 = 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-2) = 1 + 8 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{9}}{2 * 1} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{9}}{2 * 1} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2$ − не является решением, так как x ≠ −2.
Ответ: 1

$\frac{2}{x^2 - 2x + 1} - \frac{1}{x^3 - 1} = \frac{3}{x^2 + x + 1}$
$\frac{2}{(x - 1)^2} - \frac{1}{(x - 1)(x^2 + x + 1)} - \frac{3}{x^2 + x + 1} = 0$
$x^3 - 1 ≠ 0$
$x^3 ≠ 1$
x ≠ 1
$\frac{2}{(x - 1)^2} - \frac{1}{(x - 1)(x^2 + x + 1)} - \frac{3}{x^2 + x + 1} = 0$ | * $(x - 1)^2(x^2 + x + 1)$
$2(x^2 + x + 1) - (x - 1) - 3(x - 1)^2 = 0$
$2x^2 + 2x + 2 - x + 1 - 3(x^2 - 2x + 1) = 0$
$2x^2 + x + 3 - 3x^2 + 6x - 3 = 0$
$-x^2 + 7x = 0$
−x(x − 7) = 0
−x = 0
x = 0
или
x − 7 = 0
x = 7
Ответ: 0 и 7