Help SOS very urgent! помогите решить уравнения СРОЧНО!!!

Question 1

Question 2

5)
$\frac{x + 1}{x - 4} = \frac{3x + 1}{3x - 1} \\ \frac{x + 1}{x - 4} - \frac{3x + 1}{3x - 1} = 0 \\ \frac{(x + 1)(3x - 1)}{(x - 4)(3x - 1)} - \frac{(3x + 1)(x - 4)}{(x - 4)(3x - 1)} = 0 \\ \frac{3 {x}^{2} + 3x - x - 1 }{(x - 4)(3x - 1)} - \frac{3 {x}^{2} + x - 12x - 4}{( x - 4)(3x - 1) } = 0 \\ \frac{3 {x}^{2} + 2x - 1 }{(x - 4)(3x - 1)} - \frac{3 {x}^{2} - 11x - 4 }{(x - 4)(3x - 1)} = 0 \\ \frac{3 {x}^{2} + 2x - 1 - 3 {x}^{2} + 11x + 4}{(x - 4)(3x - 1)} = 0 \\ 13x + 3 = 0 \\ x = - \frac{3}{13}$
6)
$\frac{9x - 7}{3x - 2} - \frac{4x -5 }{2x - 3} = 1 \\ \frac{(9x - 7)(2x - 3) - (4x - 5)(3x - 2) - (3x - 2)(2x - 3)}{(3x - 2)(2x - 3)} = 0 \\ \frac{18 {x}^{2} - 14x - 27x + 21 - 12 {x}^{2} + 15x + 8x - 10 - 6 {x}^{2} + 4x + 9x - 6 }{(3x - 2)(2x - 3)} = 0 \\ \frac{ - 5x + 5}{(2x - 3)(3x - 2)} = 0 \\ - 5x + 5 = 0 \\ x = - 1$
7)
$\frac{2x - 1}{2x + 1} = \frac{2x + 1}{2x - 1} + \frac{4}{1 - 4 {x}^{2} } \\ \frac{2x - 1}{2x + 1} - \frac{2x + 1}{2x - 1} - \frac{4}{1 - 4 {x}^{2} } = 0 \\ \frac{2x - 1}{2x + 1} - \frac{2x + 1}{2x - 1} + \frac{4}{4 {x}^{2} - 1 } = 0 \\ \frac{ {(2x - 1)}^{2} }{(2x + 1)(2x - 1)} - \frac{ {(2x + 1)}^{2} }{(2x + 1)(2x - 1)} + \frac{4}{(2x - 1)(2x + 1)} = 0 \\ \frac{(2x - 1 + 2x + 1)(2x - 1 - 2x - 1) + 4}{(2x - 1)(2x + 1)} = 0 \\ \frac{4x \times ( - 2) + 4}{(2x - 1)(2x + 1)} = 0 \\ \frac{4( - 2x + 1)}{(2x - 1)(2x + 1)} = 0 \\ - 2x + 1 = 0 \\ x = \frac{1}{2}$
но по одз: 2х - 1 не равно 0, х не равно 1/2, значит, нет корней.

8)
$\frac{6}{ {x}^{2} + x } - \frac{x - 6}{ {x}^{2} - x } + \frac{10}{ {x}^{2} - 1} = 0 \\ \frac{6}{x(x + 1)} - \frac{x - 6}{x(x - 1)} + \frac{10}{(x - 1)(x + 1)} = 0 \\ \frac{6(x - 1)}{x(x + 1)(x - 1)} - \frac{(x - 6)(x + 1)}{x(x - 1)(x + 1)} + \frac{10x}{x(x - 1)(x + 1)} = 0 \\ \frac{6x - 6 - {x}^{2} + 6x - x + 6 + 10x }{x(x - 1)(x + 1)} = 0 \\ \frac{ - {x}^{2} + 21x}{x(x - 1)(x + 1)} = 0 \\ - {x}^{2} + 21x = 0 \\ {x}^{2} - 21x = 0 \\ x(x - 21) = 0 \\ x = 21$
х = 0 не удовл.
Ответ: 21.