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3-sqrt(x)/9-x. how do you break down?

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\[\frac{3-\sqrt x}{9-x}\] factorize the denominator use \[(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2\] cancel common factors
yes i know that but how do you that for sqrt?
use \(x=\sqrt x^2\)

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Other answers:

mmm so 3-sqet(x)=(x^2-3^9)?
your looking at the numerator (bottom bit of the fraction) you should be looking at the denominator ( the bottom bit of the fraction) \[9-x=(\cdots)(\cdots)\]
9-x...... sqert(3)-(x)??
\[9-x=3^2-\sqrt x^2=(\cdots)(\cdots)\]
(3-sq\[\left( 3-\sqrt{x} \right)\]
(3-sq\[\left( 3-\sqrt{x} \right)\](3-sq\[\left( 3-\sqrt{x} \right)\]?
\[\left( 3-\sqrt{x} \right)\]
\left( 3-sqrt{x} \right) \left( 3-sqrt{x} \right)
(3-sqrt(x)) (3-sqrt(x)
not quite\[a^2-b^2=(a+b)(a-b)\] so\[9-x=3^2-\sqrt x^2=\]
hope its right
\[9-x=3^2-\sqrt x^2=(3+\sqrt x)(3-\sqrt x)\]
so\[\frac{3-\sqrt x}{9-x}=\frac{3-\sqrt x}{(3+\sqrt x)(3-{\sqrt x)}}\]\[\qquad\qquad =\frac{\cancel{3-\sqrt x}}{(3+\sqrt x)\cancel{(3-{\sqrt x)}}}\]
what about x^3+64/x+4
I got (x+4)(x^2+4x+16)/(x+4)?
almost \[(a^3+b^3)=(a+b)(a^2-ab+b^2)\]
once you have fixed the sign error, cancel the common factor
Thank you it was very helpful!!

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