Question

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

2) 7//(X^2+4X+3)=X^2+4X+3

Answer 1

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

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

Answer 2

Большое спасибо за Ваш ответ, а 2 уравнение?