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Which choice is equivalent to the fraction below when [tex][tex]$x \geq 2$[/tex][/tex]?

Rationalize the denominator and simplify:

[tex]\frac{4}{\sqrt{x-2}-\sqrt{x}}[/tex]

A. [tex]-2(\sqrt{x}-\sqrt{x-2})[/tex]

B. [tex]2(\sqrt{x}+\sqrt{x-2})[/tex]

C. [tex]-2(\sqrt{x}+\sqrt{x-2})[/tex]

D. [tex]2(\sqrt{x}-\sqrt{x-2})[/tex]


Sagot :

Let's solve the problem step-by-step.

We start with the given fraction:
[tex]\[ \frac{4}{\sqrt{x-2} - \sqrt{x}} \][/tex]

To simplify this fraction, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(\sqrt{x-2} - \sqrt{x}\)[/tex] is [tex]\(\sqrt{x-2} + \sqrt{x}\)[/tex].

Thus, we proceed as follows:
[tex]\[ \frac{4}{\sqrt{x-2} - \sqrt{x}} \times \frac{\sqrt{x-2} + \sqrt{x}}{\sqrt{x-2} + \sqrt{x}} = \frac{4 (\sqrt{x-2} + \sqrt{x})}{(\sqrt{x-2} - \sqrt{x})(\sqrt{x-2} + \sqrt{x})} \][/tex]

Next, we simplify the denominator using the difference of squares:
[tex]\[ (\sqrt{x-2} - \sqrt{x})(\sqrt{x-2} + \sqrt{x}) = (\sqrt{x-2})^2 - (\sqrt{x})^2 = (x-2) - x = -2 \][/tex]

So, the fraction becomes:
[tex]\[ \frac{4 (\sqrt{x-2} + \sqrt{x})}{-2} \][/tex]

This simplifies by dividing both the numerator and the denominator by [tex]\(-2\)[/tex]:
[tex]\[ \frac{4 (\sqrt{x-2} + \sqrt{x})}{-2} = -2 (\sqrt{x-2} + \sqrt{x}) \][/tex]

Now we compare this result with the given choices:

A. [tex]\(-2 (\sqrt{x} - \sqrt{x-2})\)[/tex]

B. [tex]\(2 (\sqrt{x} + \sqrt{x-2})\)[/tex]

C. [tex]\(-2 (\sqrt{x} + \sqrt{x-2})\)[/tex]

D. [tex]\(2 (\sqrt{x} - \sqrt{x-2})\)[/tex]

From our simplification, we have [tex]\(-2 (\sqrt{x-2} + \sqrt{x})\)[/tex], which matches choice C.

Therefore, the correct answer is:
[tex]\[ \boxed{C} \][/tex]