Is this operation legal? Is this operation allowed?
Going from this: 
$\left ( \frac{x^{2}+6}{x^{2}-4} \right )^{2}= \left ( \frac{5x}{4-x^{2}} \right )^{2}$
To this: $\left ( \frac{\left (x^{2}+6  \right )\left ( 4-x^{2} \right )}{\left (x^{2}-4  \right )5x} \right )^{2}= 1$
 A: Assuming $x \neq 0$, then yes. Otherwise you're dividing by zero.
Hope that helps,
A: Notice that the denominators are equal, save for a factor of $-1$:  which has no impact since the fraction is squared. $$\; (4 - x^2)^2 = (-(x^2 - 4))^2 = (x^2 - 4)^2$$
So multiply both sides by $(x^2 - 4)^2$, and you'll cancel both denominators. Then there's no need to divide by $5x$. 
After multiplying both sides by $(x^2 - 4)^2$, we get
$$\left (x^{2}+6\right )^{2}= \left ( 5x \right )^{2}\iff (x^4 + 12x^2 + 36) - 25x^2 = 0 \iff x^4 - 13x^2 + 36 = (x^2 - 9)(x^2 - 4) = (x+3)(x-3)(x+2)(x-2) = 0  $$
But we need to throw out the solutions $x = 2, -2$ because the equation is not defined there.
A: Yes.
In response to your edit: You squared both sides of the equation to correct your post. However, this doesn't make a difference, it's still valid. Since you squared both sides of the equation (applied the same operation), you didn't effectively change anything.
The solution is therefore:


*

*take the square root

*multiply both sides by $(x^{2}-4)$ and divide both sides by $5x$.

*square both sides ( $1^2=1$ )
A: Yes, that operation is legal. It just comes from multiplying $\frac{(4-x^2)}{5x}$ on both sides.
A: If you are solving for $x$, it's OK as long as in the end you check whether the expression you divided by, $5x$, is non-zero.
In general, you would have to follow through with everything assuming all transformations are legal, and then check for weak spots like division by zero ($5x=0$ or $x^2-4=0$, or square roots or logs of negative numbers), and modify your set of solutions accordingly by excluding some solutions that would lead to a log of a negative number, or adding some solutions that could have lead to a zero on both sides of the equation.
A: As others have noted, it is valid so long as $x\ne0,$ since otherwise you're dividing by $0$. Let me offer another approach, using the difference of squares formula $a^2-b^2=(a-b)(a+b)$, together with the fact that $(-c)^2=c^2.$ The following, then, are equivalent:
$$\left(\frac{x^2+6}{x^2-4}\right)^2=\left(\frac{5x}{4-x^2}\right)^2$$
$$\left(\frac{x^2+6}{x^2-4}\right)^2=\left(\frac{5x}{x^2-4}\right)^2$$
$$\left(\frac{x^2+6}{x^2-4}\right)^2-\left(\frac{5x}{x^2-4}\right)^2=0$$
$$\left(\frac{x^2+6}{x^2-4}-\frac{5x}{x^2-4}\right)\left(\frac{x^2+6}{x^2-4}+\frac{5x}{x^2-4}\right)=0$$
$$\frac{x^2-5x+6}{x^2-4}\cdot\frac{x^2+5x+6}{x^2-4}=0.$$
Now, note that none of the above equations make sense when $x=\pm 2,$ since $x^2-4=(x-2)(x+2).$ So, we must assume that $x\neq 2$ and $x\neq-2$. At that point, the last equation is readily equivalent to $$(x^2-5x+6)(x^2+5x+6)=0,$$ which is true if and only if $x^2-5x+6=0$ or $x^2+5x+6=0$. These two quadratic equations give the solutions $x=\pm 2,\pm3,$ and so since we had to rule out $x=\pm 2,$ we get $x=\pm 3$.
