Question

помогите решить уравнение!
x2-6x-9/x=x2-4x-9/x2-6x-9

Answer 1

Объяснение:

$\frac{x^2-6x-9}{x}=\frac{x^2-4x-9}{x^2-6x-9}\; \; ,\; \; \; \; ODZ:\; \left \{ {{x\ne 0} \atop {x^2-6x-9\ne 0}} \right.\\\\(x^2-6x-9)^2=x\cdot (x^2-4x-9)\; \Big |\cdot \frac{1}{x^2}\; \; (x\ne 0)\\\\\frac{x^2-6x-9}{x}\cdot \frac{x^2-6x-9}{x}=\frac{x(x^2-4x-9)}{x^2}\\\\\Big (x-6-\frac{9}{x}\Big )^2=x-4-\frac{9}{x}\\\\t=x-6-\frac{9}{x}\; \; \to \; \; \; x-4-\frac{9}{x}=(x-6-\frac{9}{x})+2=t+2\\\\t^2=t+2\; \; \; \to \; \; \; t^2-t-2=0\; ,\; \; t_1=-1\; ,\; \; t_2=2\\\\\\a)\; \; x-6-\frac{9}{x}=-1\; ,\; \; \frac{x^2-6x-9}{x}=-1\; \; ,\; \; \frac{x^2-6x-9+x}{x}=0\; ,\; \frac{x^2-5x-9}{x}=0$

$x^2-5x-9=0\; \; (x\ne 0)\; ,\; \; D=25+36=61\; ,\; \; x_{1,2}=\frac{5\pm \sqrt{61}}{2}\\\\\\b)\; \; x-6-\frac{9}{x}=2\; ,\; \; \frac{x^2-6x-9}{x}=2\; ,\; \; \frac{x^2-6x-9-2x}{x}=0\; ,\; \; \frac{x^2-8x-9}{x}=0\\\\x^2-8x-9=0\; \; (x\ne 0)\; \; \to \; \; x_1=-1\; ,\; x_2=9\; \; (teorema\; Vieta)\\\\\\Otvet:\; \; x_1=-1\; ,\; x_2=9\; ,\; x_3=\frac{5-\sqrt{61}}{2}\; ,\; x_4=\frac{5+\sqrt{61}}{2}\; .$