Решите уравнение

$\frac{x(x+3)+(x+1)(x-1)}{(x-1)(x+3)} = \frac{1}{ x^{2} +2x-3}$
$\frac{ x^{2} +3x+ x^{2}-1 }{ x^{2} +3x-x-3} = \frac{1}{ x^{2} +2x-3}$
$\frac{2 x^{2} +3x-1}{ x^{2} +2x-3} = \frac{1}{ x^{2} +2x-3}$
2х²+3х-1=1
2х²+3х-2=0

D=b²-4ac=3²-4·2·(-2)=9+16=25

$x= \frac{-b+- \sqrt{D} }{ 2a}$
$x_{1} = \frac{-3+ \sqrt{25} }{ 2*2} = \frac{-3+5}{4} = \frac{2}{4} = \frac{1}{2}$
$x_{2} = \frac{-3- \sqrt{25} }{ 2*2} = \frac{-3-5}{4}= - \frac{8}{4} =-2$