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\[\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Is that the problem?

Yes!

In the denominator of the first fraction: is it -1or +1

well for one u can cross out the equation on the right, the top and bottom r exactly the same

Shoot.. you are correct. "-1"

So the denominator is 3-x-1or 2-x

now let's do the numerator multiplication:

How did you get that? :/ I'm not disagreeing, I just don't know how you worked it out..

\[(\sqrt{6-x}-2)(\sqrt{3-x}+1)=\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2\]

What is (a-b)(a+b)

(18 - 9x + x) :)

Where did that come from?

Isn't that what'll be under the radical, in the first term in the numerator? :/

It would be 18-9x+x^2

Dude, I'm tired.. my bad again.

so put that numerator over 2-x and you are done.

Can we cancel out anything else?

Are there any common factors?

|dw:1442976045683:dw|

Is this all correct, and the final version?

yep

|dw:1442976240464:dw|

This is totally wrong, right?

it is correct.

The second version is correct as well??