Question

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

Answer 1

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.