Answer
$\dfrac{x - 2}{x^2-x}$
Restrictions: $x \ne -1, 1, 0, -2$
Work Step by Step
Let's first take a look at the problem to see what we can cancel out from the numerator and denominator:
$\dfrac{x^2 - 4}{x^2 - 1} \cdot \frac{x + 1}{x^2 + 2x}$
In the first rational eexpression, we can factor the binomials in the numerator and denominator according to the formula for factoring the difference of two squares:
$a^2 - b^2 = (a + b)(a - b)$
Let's factor the numerator and denominator in the first rational expression using this formula:
$\dfrac{(x - 2)(x + 2)}{(x - 1)(x + 1)} \cdot \dfrac{x + 1}{x^2 + 2x}$
Factor $x$ from the denominator of the second expression:
$\dfrac{(x - 2)(x + 2)}{(x - 1)(x + 1)} \cdot \dfrac{x + 1}{x(x + 2)}$
Multiply the two expressions:
$\dfrac{(x - 2)(x + 2)(x + 1)}{(x - 1)(x + 1)(x)(x + 2)}$
We can cancel out $x + 2$, $x + 1$ from the expression:
$\dfrac{x - 2}{x(x - 1)}$
To find out if there are any restrictions on the variables, we need to find which values of $x$ will cause the denominator to equal zero, which would make the whole expression undefined. Let's set the denominator equal to zero and solve for $x$:
First factor:
$x^2 - 1 = 0$
Add $1$ to each side of the equation:
$x^2 = 1$
Take the square root of $1$:
$x = \pm 1$
Second factor:
$x^2 + 2x = 0$
Factor out $x$:
$x(x + 2) = 0$
Set each factor equal to zero:
$x = 0$
$x + 2 = 0$
Subtract $2$ from each side of the equation:
$x = -2$
Restrictions: $x \ne -1, 1, 0, -2$
