Question

2.13. Решите уравнение: алгебра 8 класс номер 2.13 ​

Answer 1

1) $\frac{x - 4}{x + 3} = \frac{2x + 4}{x - 3}$

ОДЗ:

$x + 3 ≠0 \\ x≠ - 3$,

$x - 3≠0 \\ x≠3$;

$(x - 4)(x - 3) = (x + 3)(2x + 4)$

${x}^{2} - 3x - 4x + 12 = 2 {x}^{2} + 4x + 6x + 12$

${x}^{2} - 7x + 12 = 2 {x}^{2} + 10x + 12$

${x}^{2} - 2 {x}^{2} - 7x - 10x + 12 - 12 = 0$

$- {x}^{2} - 17x = 0$

${x}^{2} + 17x = 0$

$x(x + 17) = 0$

$x_1 = 0 \: \: \: \: x + 17 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_2 = - 17$

Ответ: $x = 0$; $x = - 17$

2) $\frac{2 - 3y}{3 - y} = \frac{y + 2}{y + 3}$

ОДЗ:

$3 - y≠0 \\ y≠3$,

$y + 3≠0 \\ y≠ - 3$;

$(2 - 3y)(y + 3) = (3 - y)(y + 2)$

$2y + 6 - 3 {y}^{2} - 9y = 3y + 6 - {y}^{2} - 2y$

$- 3 {y}^{2} - 7y + 6 = - {y}^{2} + y + 6$

$- 3 {y}^{2} + {y}^{2} - 7y - y + 6 - 6 = 0$

$- 2 {y}^{2} - 8y = 0$

${y}^{2} + 4y = 0$

$y(y + 4) = 0$

$y_1 = 0 \: \: \: \: \: y + 4 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y_2 = - 4$

Ответ: $y = 0$; $y = - 4$

3) $\frac{x - 1}{x + 1} + \frac{x + 2}{x - 1} + 3 = 0$

ОДЗ:

$x + 1≠0 \\ x≠ - 1$,

$x - 1≠0 \\ x≠1$;

$\frac{(x - 1) \times (x - 1)}{(x + 1) \times (x - 1)} + \frac{(x + 2) \times (x + 1)}{(x - 1) \times (x + 1)} + \frac{3(x + 1)(x - 1)}{(x + 1)(x - 1)} = 0$

$\frac{(x - 1)(x - 1) + (x + 2)(x + 1) + 3(x + 1)(x - 1)}{(x + 1)(x - 1)} = 0$

$\frac{ {x}^{2} - 2x + 1 + {x}^{2} + 3x + 2 + 3 {x}^{2} - 3 }{(x + 1)(x - 1)} = 0$

$\frac{5 {x}^{2} + x}{(x + 1)(x - 1)} = 0$

$5 {x}^{2} + x = 0$

$x(5x + 1) = 0$

$x_1 = 0 \: \: \: \: \: 5x + 1 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_2 = - 0.2$

Ответ: $x = 0$; $x = - 0.2$

4) $\frac{7x + 5}{x - 1} - \frac{2x}{x + 1} = - 5$

$\frac{7x + 5}{x - 1} - \frac{2x}{x + 1} + 5 = 0$

ОДЗ:

$x - 1≠0 \\ x≠1$,

$x + 1≠0 \\ x≠ - 1$;

$\frac{(7x + 5) \times (x + 1)}{(x - 1) \times (x + 1)} - \frac{2x \times (x - 1)}{(x + 1) \times (x - 1)} + \frac{5(x - 1)(x + 1)}{(x - 1)(x + 1)} = 0$

$\frac{(7x + 5)(x + 1) - 2x(x - 1) + 5(x - 1)(x + 1)}{(x - 1)(x + 1)} = 0$

$\frac{7 {x}^{2} + 7x + 5x + 5 - 2 {x}^{2} + 2x + 5 {x}^{2} - 5 }{(x - 1)(x + 1)} = 0$

$\frac{10 {x}^{2} + 14x}{(x - 1)(x + 1)} = 0$

$10 {x}^{2} + 14x = 0$

$5 {x}^{2} + 7x = 0$

$x(5x + 7) = 0$

$x_1 = 0 \: \: \: \: \: 5x + 7 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_2 = - 1.4$

Ответ: $x = 0$; $x = - 1.4$