7) x^2 + 4x / (x - 5) - (9x + 50) / (x - 5) = 0

Let's solve these equations step by step. Remember that the denominator cannot be zero, so we must note the restrictions first. 7) $\frac{x^2 + 4x}{x - 5} - \frac{9x + 50}{x - 5} = 0$ Restriction: $x \neq 5$. Since denominators are the same, combine the numerators: $x^2 + 4x - (9x + 50) = 0$ $x^2 + 4x - 9x - 50 = 0$ $x^2 - 5x - 50 = 0$ Solving via discriminant ($D = b^2 - 4ac$): $D = (-5)^2 - 4 \cdot 1 \cdot (-50) = 25 + 200 = 225 = 15^2$ $x_1 = \frac{5 + 15}{2} = 10$ $x_2 = \frac{5 - 15}{2} = -5$ Both values satisfy $x \neq 5$. **Ответ: -5; 10.** 8) $\frac{x^2 - 6x}{x - 3} + \frac{15 - 2x}{x - 3} = 0$ Restriction: $x \neq 3$. Combine numerators: $x^2 - 6x + 15 - 2x = 0$ $x^2 - 8x + 15 = 0$ Using Vieta's formulas ($x_1 + x_2 = 8$, $x_1 \cdot x_2 = 15$): $x_1 = 3$, $x_2 = 5$ Since the restriction is $x \neq 3$, the root $x = 3$ is extraneous. **Ответ: 5.** 9) $\frac{x^2 - 6x}{x - 4} = 4$ Restriction: $x \neq 4$. Multiply both sides by $(x - 4)$: $x^2 - 6x = 4(x - 4)$ $x^2 - 6x = 4x - 16$ $x^2 - 10x + 16 = 0$ Using Vieta's formulas ($x_1 + x_2 = 10$, $x_1 \cdot x_2 = 16$): $x_1 = 2$, $x_2 = 8$ Both values satisfy $x \neq 4$. **Ответ: 2; 8.**