#### Answer

Conditional equation.
$x=2$

#### Work Step by Step

Given equation,
$\frac{x+2}{x+3} + \frac{1}{x^{2}+2x-3} - 1 = 0$
By factoring, $x^{2}+2x-3=(x+3)(x-1)$,
the given equation can be written as
$\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} - 1 = 0$
$\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1$
$x=1 $ or $ x=-3$ makes the denominator zero, so,
$\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$
Taking LCD,
$ \frac{(x+2)(x-1)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$
$ \frac{(x^{2}+2x-x-2)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$
$ \frac{x^{2}+x-1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$
$ x^{2}+x-1 = (x+3)(x-1) ; x\ne 1,-3;$
$ x^{2}+x-1 = x^{2}+2x-3 ; x\ne 1,-3;$
$ x^{2}+x-1 -x^{2}-2x+3=0 ; x\ne 1,-3;$
$-x+2=0 ; x\ne 1,-3;$
$x=2 ; x\ne 1,-3;$
This equation has one solution, so it is a conditional equation.