The given fraction is equivalent to [tex]3(\sqrt{x}+\sqrt{x-3})[/tex]. Then the correct answer is option C.
What is rationalization?
The elimination of radicals from an algebraic fraction's denominator is known as root rationalization in elementary algebra.
Rationalizing the given expression to get the solution.
[tex]\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9}{\sqrt{x}-\sqrt{x-3}}\times \dfrac{\sqrt{x}+\sqrt{x-3}}{\sqrt{x}+\sqrt{x-3}}[/tex]
[tex]\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{(\sqrt{x})^2-(\sqrt{x-3})^2}[/tex]
[tex]\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{({x})-({x-3})}[/tex]
[tex]\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{3}={3(\sqrt{x}-\sqrt{x-3}[/tex]
Hence, the correct answer is option C.
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