Simplifying expressions with radical exponents

Question

This is the expression that needs to be simplified:

$\frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$

My first step is to move the negative exponent to the denominator:

$\frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$

Then multiply the denominator and add the exponents:

$\frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$

I don't know the next steps to reach the given answer of:

$\frac{9}{(9-x^2)^{3/2}}$

Answer 1

Hint: Write $$\frac{(9-x^2)^{1/2}}{9-x^2}+\frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is $$\frac{1}{\sqrt{9-x^2}}+\frac{x^2}{(9-x^2)^{3/2}}$$

Answer 2

Your first step is incorrect.

$$\frac{a+b^{-n}}{c} \color{red}{\neq \frac{a}{b^n+c}}$$

You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in

$$\frac{ab^{-n}}{c} = \frac{a}{b^nc}$$

Instead, you can make the substitution $u = 9-x^2$:

$$\frac{u^{\frac{1}{2}} + x^2(u)^{-\frac{1}{2}}}{u} = \frac{u^{\frac{1}{2}}(1+x^2u^{-1})}{u} = \frac{1+x^2u^{-1}}{u^{\frac{1}{2}}} = \frac{1}{u^{\frac{1}{2}}}+\frac{x^2u^{-1}}{u^{\frac{1}{2}}} = \frac{1}{u^{\frac{1}{2}}}+\frac{x^2}{u\cdot u^{\frac{1}{2}}}$$

Plugging in $u = 9-x^2$ and simplifying gives

$$\frac{1}{(9-x^2)^{\frac{1}{2}}}+\frac{x^2}{(9-x^2)(9-x^2)^{\frac{1}{2}}} = \frac{9-x^2}{(9-x^2)(9-x^2)^{\frac{1}{2}}}+\frac{x^2}{(9-x^2)(9-x^2)^{\frac{1}{2}}} = \frac{9}{(9-x^2)^{\frac{3}{2}}}$$

Answer 3

Consider the substitution $z = \sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.