Помогите решить уравнения: (Дам 15 баллов за лучший ответ!!!) 1) X//(X+3)-(X-4)//(3-X)=1...

Решите задачу:

$\frac{x}{x+3} - \frac{x-4}{3-x} = 1, \\ \left \{ {{x+3 \neq 0,} \atop {x-3 \neq 0;}} \right. \left \{ {{x \neq -3,} \atop {x \neq 3;}} \right. \\ \frac{x}{x+3} + \frac{x-4}{x-3} - 1 = 0, \\ \frac{x(x-3)+(x-4)(x+3)-(x+3)(x-3)}{(x+3)(x-3)} = 0, \\ \frac{x^2-3x+x^2+3x-4x-12-(x^2-9)}{x^2-9} = 0, \\ \frac{2x^2-4x-12-x^2+9}{x^2-9} = 0, \\ \frac{x^2-4x-3}{x^2-9} = 0, \\ x^2-4x-3=0, \\ D=7, \\ x_1=2- \sqrt{7}, x_2=2+ \sqrt{7}.$

$\frac{7}{x^2+4x+3}=x^2+4x+3, \\ x^2+4x+3 \neq 0, \\ \left \{ {{x \neq -3,} \atop {x \neq -1;}} \right. \\ x^2+4x+3=t, \\ \frac{7}{t}=t, \\ \frac{7}{t}-t=0, \\ \frac{7-t^2}{t}=0, 7-t^2=0, \\ t^2=7, \\ t_1=- \sqrt{7}, t_2= \sqrt{7}, \\ \left [ {{x^2+4x+3=- \sqrt{7},} \atop {x^2+4x+3= \sqrt{7};}} \right. \left [ {{x^2+4x+3+ \sqrt{7}=0,} \atop {x^2+4x+3-\sqrt{7}=0;}} \right. \\$
$x^2+4x+3+ \sqrt{7}=0, \\ D=4-(3+ \sqrt{7})=1- \sqrt{7}\approx-1,6<0; \\ x\in\varnothing; \\ x^2+4x+3-\sqrt{7}=0, \\ D=4-(3-\sqrt{7})=1+\sqrt{7}, \\ x_1=-2- \sqrt{1+\sqrt{7}}, x_2=-2+\sqrt{1+\sqrt{7}}.$