anonymous
  • anonymous
3-sqrt(x)/9-x. how do you break down?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
\[\frac{3-\sqrt x}{9-x}\] factorize the denominator use \[(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2\] cancel common factors
anonymous
  • anonymous
yes i know that but how do you that for sqrt?
UnkleRhaukus
  • UnkleRhaukus
use \(x=\sqrt x^2\)

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More answers

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

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