3) a) 5/(x-2) + 1 = 14/(x²-4x+4);

3) a) Решим уравнение:$$\frac{5}{x-2} + 1 = \frac{14}{x^2-4x+4}$$ОДЗ: $$x
eq 2$$$$\frac{5}{x-2} + 1 = \frac{14}{(x-2)^2}$$$$\frac{5 + x-2}{x-2} = \frac{14}{(x-2)^2}$$$$\frac{x+3}{x-2} = \frac{14}{(x-2)^2}$$Умножим обе части на $$(x-2)^2$$:$$x^2 + x - 6 = 14$$$$x^2 + x - 20 = 0$$$$D = 1 - 4(-20) = 1+80 = 81$$$$x_1 = \frac{-1 + 9}{2} = 4$$$$x_2 = \frac{-1 - 9}{2} = -5$$Ответ: x = 4, x = -5