Помогите по алгебре. даю 50 баллов. 8 класс. Тема: Квадратный трехчлен. Решение...

$\frac{6x-10}{x^{2}-4}=\frac{3x-2x^{2}}{4-x^{2}}\\\frac{6x-10}{x^{2}-4} - \frac{3x-2x^{2}}{4-x^{2}} = 0\\ \frac{6x-10}{x^{2}-4} + \frac{3x-2x^{2}}{x^{2} - 4} = 0\\ \\ {x}^{2} - 4≠0 \\ {x}^{2}≠4 \\ x≠ - 2 \\ x≠ 2$

$6x - 10 + 3x - 2 {x}^{2} = 0 \\ - 2 {x}^{2} + 9x - 10 = 0 \\ 2 {x}^{2} - 9x + 10 = 0 \\ D = 81 - 4 \times 2 \times 10 = 1 \\ x_{1} = \frac{9 - 1}{2 \times 2} = \frac{8}{4} = 2 \\ x_{2} = \frac{9 + 1}{2 \times 2} = \frac{10}{4} = 2.5 \\$

2 - посторонний корень

Ответ: 2,5

$\frac{3x}{x-2}+\frac{18}{x^{2}-2x}+\frac{7}{x}=0 \\ \frac{3x}{x-2}+\frac{18}{x(x - 2)}+\frac{7}{x}=0 \\ \frac{3x \times x + 18 + 7(x - 2)}{x(x - 2)}=0 \\ \frac{3 {x}^{2} + 18 + 7x - 14}{x(x - 2)}=0 \\ \frac{3 {x}^{2} + 7x + 4}{x(x - 2)}=0 \\ \\ x(x - 2)≠0 \\ x≠0 \\ x - 2≠0 \\ x≠2 \\ \\ 3 {x}^{2} + 7x + 4 = 0 \\ D = 49 - 4 \times 3 \times 4 = 1 \\ \\ x _{1} = \frac{ - 7 - 1}{2 \times 3} = \frac{ - 8}{6} = \\ = - \frac{4}{3} = - 1 \frac{1}{3} \\ x _{2} = \frac{ - 7 + 1}{2 \times 3} = \frac{ - 6}{6} = - 1$

Ответ:

$(- 1 \frac{1}{3}; - 1)$