Answer
$\frac{(x+1)}{(x-3)}$.
Work Step by Step
The given expression is
$=\frac{x^2-1}{x^2}\div \frac{x^2-4x+3}{x^2}$
$=\frac{x^2-1}{x^2}\times \frac{x^2}{x^2-4x+3}$
Cancel common terms.
$=\frac{x^2-1}{x^2-4x+3}$
Factor the numerator and denominator.
Numerator $=x^2-1$.
$=x^2-1^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$
$=(x+1)(x-1)$.
Denominator $=x^2-4x+3$.
Rewrite the middle term $-4x$ as $-3x-x$.
$=x^2-3x-x+3$
Group terms.
$=(x^2-3x)+(-x+3)$
Factor each term.
$=x(x-3)-1(x-3)$
Factor out $(x-3)$.
$=(x-3)(x-1)$
Substitute factors.
$=\frac{(x+1)(x-1)}{(x-3)(x-1)}$
Cancel common terms.
$=\frac{(x+1)}{(x-3)}$.
