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\[\frac x{x^2 - 4} \ge 0\]
???

yes
that is the problem

cross multiply..what do you get?

1x=0

= ?

where did you get =?

1x = 0

should be \(\ge\) not =

your initial sign was \(\ge\) so you can't change it

right

so my answer is correct ?
or is there more to do ?

nothing more

or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or -2

Uh... @lgbasallote , that answer isn't right.

you sure?

yep:)

then why don't you show how it's done, oh great one?

It's \(x\in (-2, 0]\cup (2,\infty)\) Well, that's what I'm getting... let me check.

i did say x cannot be 2 or -2

you did read what i was saying right?

Yes, but, it's supposed to be:
\[
x>2\\
-2

I'd look at where x^3-4x is positive and negative...

simplest way I think...

i didn't say any signs did i? i said x cannot be 2 or -2

So how does one get to this conclusion? If you dont mind me asking.

and i was also insulting the mathematicians

multiply both sides by (x^2-4 )^2

Here's a counter example to your solution:
\[
x=1\ge0\\
\frac{1}{1-4}=-\frac{1}{3}\not \ge\; 0
\]

counter example to what?

(@lgbasallote 's not yours, @satellite73 , haha)

oh, ok

oh

anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine

if it's a computer you're trying to please

Thank you. The other question I have goes like this : a^5/3 * b^2/3 * c^3 / C^-2 * a^2/3 * b^-4/3 .

add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)

for example on (a) would leave denominator 3 alone and just add the numnerators?

Hope you're not asking about adding fractions :P

Nope!

My answer is A^1 * b^6/3 * c^5

would this be correct?