Помогите пожалуйста решить

Решите задачу:

$1)\quad x \leq 7- \frac{16}{x+1} \; ,\; \; \; \; ODZ:\; x\ne -1\\\\x-7+\frac{16}{x+1} \leq 0\\\\ \frac{(x-7)(x+1)+16}{x1} \leq 0\\\\ \frac{x^2-6x+9}{x+1} \leq 0\\\\ \frac{(x-3)^2}{x+1} \leq 0\quad ---(-1)+++[3]+++\\\\x\in (-\infty ,-1)U\{3\}$

$2)\quad \frac{1}{2x^2-x-3} + \frac{1}{3x^2+x-2} = \frac{1}{6x^2+7x+1} \\\\ODZ:\; 2x^2-x-3\ne0,\; 3x^2+x-2\ne 0,\; 6x^2+7x+1\ne 0\; \to\\\\x\ne -1;\; x\ne \frac{3}{2};\; x\ne \frac{2}{3};\; x\ne -\frac{1}{6}\\\\ \frac{1}{2(x+1)(x-\frac{3}{2})} + \frac{1}{3(x+1)(x-\frac{2}{3})} - \frac{1}{6(x+1)(x+\frac{1}{6})} =0\\\\ \frac{(3x-2)(6x+1)+(2x-3)(6x+1)-(2x-3)(3x-2)}{(x+1)(2x-3)(3x-2)(6x+1)} =0\\\\ \frac{18x^2-9x-2+12x^2-16x-3-6x^2-13x+6}{(x+1)(2x-3)(3x-2)(6x+1)} =0\\\\ 24x^2-38x+1 =0\\\\D/4=337$

$x_1= \frac{19-\sqrt{337}}{24} \; ;\; \; x_2= \frac{19+\sqrt{337}}{24}$