Решите уравнение: а) б) в)

Решите задачу:

$1)\; \frac{x^2-3x-4}{x+1}=0\; \; \Rightarrow \; \; \left \{ {{x^2-3x-4=0} \atop {x+1\ne 0}} \right. \; \left \{ {x_1=-1,\; x_2=4} \atop {x\ne -1}} \right. \; \Rightarrow \; \underline {x=4}\\\\2)\; x+7=\frac{8}{x}\; \; \Rightarrow \; \; \frac{x^2+7x-8}{x} =0\; ,\; \left \{ {{x^2+7x-8=0} \atop {x\ne 0}} \right. \; ,\; \left \{ {{x_1=-8,\; x_2=1} \atop {x\ne 0}} \right. \; \Rightarrow \\\\\underline {x_1=-8,\; x_2=1}$

$3)\; \frac{x}{x+2} + \frac{x+2}{x-2} = \frac{8}{x^2-4} \; ,\; \; ODZ:\; \left \{ {{x+2\ne 0,\; x-2\ne 0} \atop {x^2-4\ne 0}} \right. \; \Rightarrow \; \; x\ne \pm 2\\\\ \frac{x(x-2)+(x+2)^2-8}{(x+2)(x-2)} =0\\\\ \frac{x^2-2x+x^2+4x+4-8}{(x+2)(x-2)} =0\\\\ \frac{2x^2+2x-4}{(x+2)(x-2)} =0\; ,\; \; \frac{2(x^2+x-2)}{(x+2)(x-2)} =0\; ,\; \frac{2(x-1)(x+2)}{(x+2)(x-2)} =0\; \Rightarrow \\\\\frac{2(x-1)}{x-2}=0\; \; \Rightarrow \; \; \underline {x=1}$