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Question 1 of 10

Which choice is equivalent to the fraction below when [tex]\( x \geq 3 \)[/tex]?

[tex]\[ \frac{9}{\sqrt{x}-\sqrt{x-3}} \][/tex]

A. [tex]\( -3(\sqrt{x}+\sqrt{x-3}) \)[/tex]

B. [tex]\( 3(\sqrt{x}-\sqrt{x-3}) \)[/tex]

C. [tex]\( -3(\sqrt{x}-\sqrt{x-3}) \)[/tex]

D. [tex]\( 3(\sqrt{x}+\sqrt{x-3}) \)[/tex]


Sagot :

To determine which choice is equivalent to the fraction [tex]\(\frac{9}{\sqrt{x} - \sqrt{x-3}}\)[/tex] when [tex]\(x \geq 3\)[/tex], we need to rationalize the denominator. Here is a step-by-step solution:

1. Given Fraction:
[tex]\[ \frac{9}{\sqrt{x} - \sqrt{x-3}} \][/tex]

2. Rationalize the Denominator:
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(\sqrt{x} - \sqrt{x-3}\)[/tex] is [tex]\(\sqrt{x} + \sqrt{x-3}\)[/tex].

3. Multiply Numerator and Denominator by the Conjugate:
[tex]\[ \frac{9}{\sqrt{x} - \sqrt{x-3}} \cdot \frac{\sqrt{x} + \sqrt{x-3}}{\sqrt{x} + \sqrt{x-3}} = \frac{9 \cdot (\sqrt{x} + \sqrt{x-3})}{(\sqrt{x} - \sqrt{x-3}) \cdot (\sqrt{x} + \sqrt{x-3})} \][/tex]

4. Simplify the Denominator:
Applying the difference of squares formula [tex]\((a - b)(a + b) = a^2 - b^2\)[/tex]:
[tex]\[ (\sqrt{x} - \sqrt{x-3})(\sqrt{x} + \sqrt{x-3}) = (\sqrt{x})^2 - (\sqrt{x-3})^2 = x - (x - 3) = x - x + 3 = 3 \][/tex]

5. Simplify the Fraction:
[tex]\[ \frac{9 (\sqrt{x} + \sqrt{x-3})}{3} \][/tex]

6. Reduce the Fraction:
[tex]\[ \frac{9 (\sqrt{x} + \sqrt{x-3})}{3} = 3(\sqrt{x} + \sqrt{x-3}) \][/tex]

Therefore, the simplified and rationalized form of the given fraction is:
[tex]\[ 3(\sqrt{x} + \sqrt{x-3}) \][/tex]

7. Compare with the Choices:
- A: [tex]\(-3(\sqrt{x} + \sqrt{x-3})\)[/tex]
- B: [tex]\(3(\sqrt{x} - \sqrt{x-3})\)[/tex]
- C: [tex]\(-3(\sqrt{x} - \sqrt{x-3})\)[/tex]
- D: [tex]\(3(\sqrt{x} + \sqrt{x-3})\)[/tex]

Answer:
The correct choice is:
[tex]\[ \boxed{D. \ 3(\sqrt{x}+\sqrt{x-3})} \][/tex]