simplify\:\frac{x+2}{\frac{x}{\frac{x^2-4}{x+3}}^2}

Answer 1

We can begin by simplifying the expression inside the parentheses of the denominator:

$$\frac{x^2-4}{x+3}=\frac{(x-2)(x+2)}{x+3}$$

So our denominator becomes:

$$\left(\frac{x}{\frac{(x-2)(x+2)}{x+3}}\right)^2 = \left(\frac{x+3}{x-2}\cdot\frac{x+3}{x+2}\cdot x\right)^2$$

To simplify further, we can write the expression in terms of factors:

$$\left(\frac{(x+3)^2\cdot x}{(x-2)(x+2)}\right)^2$$

Now, we can substitute all of this back into the original expression:

$$\frac{x+2}{\left(\frac{(x+3)^2\cdot x}{(x-2)(x+2)}\right)^2} = \frac{x+2}{\frac{(x+3)^4\cdot x^2}{(x-2)^2(x+2)^2}}$$

We can simplify further by dividing the numerator and denominator by $x+2$:

$$\frac{x+2}{\frac{(x+3)^4\cdot x^2}{(x-2)^2(x+2)^2}} = \frac{(x+2)^2(x-2)^2}{(x+3)^4\cdot x^2}$$

And this is our final simplified expression.