Это уравнение с дробью x______ x+1________ 1---- + ------- =...

x≠-3, x≠1

$\frac{x}{x-1}+\frac{x+1}{x+3}=\frac{1}{x^2+2x-3}\\\\\frac{x}{x-1}+\frac{x+1}{x+3}=\frac{1}{x^2+3x-x-3}\\\\\frac{x}{x-1}+\frac{x+1}{x+3}=\frac{1}{x(x+3)-(x+3)}\\\\\frac{x}{x-1}+\frac{x+1}{x+3}=\frac{1}{(x+3)(x-1)}\\\\\frac{x}{x-1}+\frac{x+1}{x+3}-\frac{1}{(x+3)(x-1)}=0\\\\\frac{x(x+3)}{(x+3)(x-1)}+\frac{(x+1)(x-1)}{{(x+3)(x-1)}}-\frac{1}{(x+3)(x-1)}=0\\\\\frac{x^2+3x+x^2-1-1}{(x+3)(x-1)}=0\\\\\frac{2x^2+3x-2}{(x+3)(x-1)}=0\\\\2x^2+3x-2=0\\\\D=3^2-4\cdot2\cdot(-2)=9+16=25\\\\\sqrt{D}=\sqrt{25}=5$

$x_1=\frac{-3-5}{2\cdot2}=\frac{-8}{4}=-2\\\\x_2=\frac{-3+5}{2\cdot2}=\frac{2}{4}=\frac{1}{2}$

Ответ: -2; 1/2