Реши уравнение: a) x/(x-3) - 5/(x+3) = 18/(x²-9)

Решим уравнения. a) $\frac{x}{x-3} - \frac{5}{x+3} = \frac{18}{x^2-9}$ \begin{aligned} &\frac{x}{x-3} - \frac{5}{x+3} = \frac{18}{(x-3)(x+3)} \\ &\frac{x(x+3) - 5(x-3)}{(x-3)(x+3)} = \frac{18}{(x-3)(x+3)} \\ &x^2 + 3x - 5x + 15 = 18 \\ &x^2 - 2x - 3 = 0 \\ D &= (-2)^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16 \\ x_1 &= \frac{2 + \sqrt{16}}{2} = \frac{2 + 4}{2} = 3 \\ x_2 &= \frac{2 - \sqrt{16}}{2} = \frac{2 - 4}{2} = -1 \end{aligned} $x_1 = 3$ не является решением, так как на него делить нельзя. **Ответ: $x = -1$** б) $\frac{70}{x^2-16} - \frac{17}{x-4} = \frac{3x}{x+4}$ \begin{aligned} &\frac{70}{(x-4)(x+4)} - \frac{17}{x-4} = \frac{3x}{x+4} \\ &\frac{70 - 17(x+4)}{(x-4)(x+4)} = \frac{3x(x-4)}{(x-4)(x+4)} \\ &70 - 17x - 68 = 3x^2 - 12x \\ &3x^2 - 12x + 17x - 2 = 0 \\ &3x^2 + 5x - 2 = 0 \\ D &= 5^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49 \\ x_1 &= \frac{-5 + \sqrt{49}}{2 \cdot 3} = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \\ x_2 &= \frac{-5 - \sqrt{49}}{2 \cdot 3} = \frac{-5 - 7}{6} = \frac{-12}{6} = -2 \end{aligned} **Ответ: $x_1 = \frac{1}{3}, x_2 = -2$** в) $\frac{3}{(2-x)^2} - \frac{5}{(x+2)^2} = \frac{14}{x^2-4}$ Домножим правую часть на -1: $\frac{3}{(2-x)^2} - \frac{5}{(x+2)^2} = -\frac{14}{(x-2)(x+2)}$ $\frac{3}{(x-2)^2} - \frac{5}{(x+2)^2} = -\frac{14}{(x-2)(x+2)}$ $\frac{3(x+2)^2 - 5(x-2)^2}{(x-2)^2(x+2)^2} = -\frac{14}{(x-2)(x+2)}$ $\frac{3(x^2+4x+4) - 5(x^2-4x+4)}{(x-2)(x+2)(x-2)(x+2)} = -\frac{14}{(x-2)(x+2)}$ $\frac{3x^2+12x+12 - 5x^2+20x-20}{(x-2)(x+2)(x-2)(x+2)} = -\frac{14}{(x-2)(x+2)}$ $\frac{-2x^2+32x-8}{(x-2)(x+2)(x-2)(x+2)} = -\frac{14}{(x-2)(x+2)}$ $-2x^2+32x-8 = -14(x-2)(x+2)$ $-2x^2+32x-8 = -14(x^2-4)$ $-2x^2+32x-8 = -14x^2+56$ $12x^2+32x-64 = 0$ $3x^2+8x-16 = 0$ $D = 8^2 - 4 \cdot 3 \cdot (-16) = 64 + 192 = 256$ $x_1 = \frac{-8 + \sqrt{256}}{2 \cdot 3} = \frac{-8 + 16}{6} = \frac{8}{6} = \frac{4}{3}$ $x_2 = \frac{-8 - \sqrt{256}}{2 \cdot 3} = \frac{-8 - 16}{6} = \frac{-24}{6} = -4$ **Ответ: $x_1 = \frac{4}{3}, x_2 = -4$** г) $\frac{2}{4-x^2} - \frac{1}{2x-4} - \frac{7}{2x^2+4x} = 0$ $\frac{2}{(2-x)(2+x)} + \frac{1}{4-2x} - \frac{7}{2x^2+4x} = 0$ $\frac{2}{(2-x)(2+x)} - \frac{1}{2(x-2)} - \frac{7}{2x(x+2)} = 0$ $\frac{-2}{-(2-x)(2+x)} - \frac{1}{2(x-2)} - \frac{7}{2x(x+2)} = 0$ $\frac{-2}{(x-2)(2+x)} - \frac{1}{2(x-2)} - \frac{7}{2x(x+2)} = 0$ $\frac{-4x-x(x+2)-14(x-2)}{2x(x-2)(x+2)} = 0$ $\frac{-4x-x^2-2x-7x+14}{2x(x-2)(x+2)} = 0$ $\frac{-x^2-13x+14}{2x(x-2)(x+2)} = 0$ $-x^2-13x+14 = 0$ $x^2+13x-14 = 0$ $D = 13^2 - 4 \cdot 1 \cdot (-14) = 169 + 56 = 225$ $x_1 = \frac{-13 + \sqrt{225}}{2 \cdot 1} = \frac{-13 + 15}{2} = \frac{2}{2} = 1$ $x_2 = \frac{-13 - \sqrt{225}}{2 \cdot 1} = \frac{-13 - 15}{2} = \frac{-28}{2} = -14$ **Ответ: $x_1 = 1, x_2 = -14$**