Помогите!!! 3 задание

Решите задачу:

$1)\quad \frac{4}{x+5} - \frac{3}{x-1} = \frac{36}{x^2+4x-5}-1\\\\x^2+4x-5=(x+5)(x-1)\\\\\frac{4(x-1)-3(x+5)-36+x^2+4x-5}{(x+5)(x-1)}=0\\\\\frac{4x-4-3x-15-36+x^2+4x-5}{(x+5)(x-1)} =0\\\\ \frac{x^2+5x-60}{(x+5)(x-1)} =0\; \; \to \; \; \left \{ {{x^2+5x-60=0} \atop {x\ne -5,\; x\ne 1}} \right. \\\\D=25+240=265\\\\Otet:\; \; x_1= \frac{-5-\sqrt{265}}{2} \; ;\; \; x_2= \frac{-5+\sqrt{265}}{2}\; .$

$2)\quad \frac{x}{x+1} + \frac{x+2}{x+6} - \frac{8-2x}{x^2+7x+6}=0\\\\x^2+7x+6=(x+1)(x+6)\\\\ \frac{x(x+6)+(x+2)(x+1)-8+2x}{(x+1)(x+6)} =0\\\\ \frac{x^2+6x+x^2+3x+2-8+2x}{(x+1)(x+6)} =0\\\\ \frac{2x^2+11x-6}{(x+1)(x+6)} =0\; \; \to \; \; \left \{ {{2x^2+11x-6=0} \atop {x\ne -1\; ,\; x\ne -6}} \right. \\\\D=121+48=169\\\\x_1= \frac{-11-13}{4} =-6\notin ODZ\; ;\; \; x_2= \frac{-11+13}{4} =\frac{1}{2}\; .\\\\Otvet:\; \; x=\frac{1}{2}\; .$