How do you simply the following expression: (((sqrt(6-x))-2)/((sqrt(3-x)-1)))*(((sqrt(3-x))+1)/((sqrt(3-x))+1))) Thank you very much! Any and all help is greatly appreciated!

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How do you simply the following expression: (((sqrt(6-x))-2)/((sqrt(3-x)-1)))*(((sqrt(3-x))+1)/((sqrt(3-x))+1))) Thank you very much! Any and all help is greatly appreciated!

Calculus1
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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!

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Other answers:

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"
If the problem is as I have posted it the denominators are of the form (a-b)(a+b) and that is equal to a^2-b^2
I know, that's the point: Multiplying by the conjugate, but I'm weary as to how to multiple that across... ?
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??
That was my initial, instinctual way to multiple the numerator across, but I thought for sure it was wrong, and I understand how you first showed me, foiling the polynomials UNDER the radical, getting 18 - 9x + x^2. But are you saying that my instinctual way works?

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