Question

2x+3/x^2-2x-x-3/x^2+2x =0

Answer 1

$dfrac{2x+3}{x^2-2x}-dfrac{x-3}{x^2+2x}=0\\dfrac{2x+3}{x(x-2)}-dfrac{x-3}{x(x+2)}=0\\\dfrac{(2x+3)(x+2)}{x(x-2)(x+2)}-dfrac{(x-3)(x-2)}{x(x+2)(x-2)}=0\\dfrac{(2x+3)(x+2)-(x-3)(x-2)}{x(x-2)(x+2)}=0\\displaystyleleft { {{(2x+3)(x+2)-(x-3)(x-2)=0} atop {boldsymbol{xneq 0;~xneq 2;~xneq -2}}} right. \\left { {{2x^2+3x+4x+6-(x^2-3x-2x+6)=0} atop {xneq 0;~xneq 2;~xneq -2}} right.\\left { {{2x^2+7x+6-x^2+5x-6=0} atop {xneq 0;~xneq 2;~xneq -2}} right.$

$displaystyleleft { {{x^2+12x=0} atop {xneq 0;~xneq 2;~xneq -2}} right.\\left { {{x(x+12)=0} atop {xneq 0;~xneq 2;~xneq -2}} right. \\left { {{x=0;~x=-12} atop {xneq 0;~xneq 2;~xneq -2}} right. \\boxed{boldsymbol{x=-12}}$