Question

using the conjugate to rationalize the denominator
use the work shown to answer the following question.
\\(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{x}}\left(\frac{\sqrt{3}+\sqrt{x}}{\sqrt{3}+\sqrt{x}}\
ight)\\)
\\(=\frac{\sqrt{3}(\sqrt{3}+\sqrt{x})}{(\sqrt{3}-\sqrt{x})(\sqrt{3}+\sqrt{x})}\\)
\\(=\frac{\sqrt{3}(\sqrt{3})+\sqrt{3}(\sqrt{x})}{(\sqrt{3})^2-(\sqrt{x})^2}\\)
what is \\(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{x}}\\) in simplest form?
options (from left to right, top to bottom? wait, the options are four vertical rectangles with fractions: first (rightmost? wait the image shows four options, lets parse the ocr for options: first option (rightmost) is \\(\frac{3 + \sqrt{3x}}{3 - x}\\), then next is \\(\frac{9 + \sqrt{3x}}{3 + x}\\), then \\(\frac{\sqrt{3} + \sqrt{3x}}{3 + x}\\), then \\(\frac{9 + \sqrt{3x}}{3 - x}\\) – wait no, lets re-express the ocr of the options: the four options (from right to left? wait the image has four vertical boxes. lets list the ocr of each options fraction:

1. \\(\frac{3 + \sqrt{3x}}{3 - x}\\)
2. \\(\frac{9 + \sqrt{3x}}{3 + x}\\)
3. \\(\frac{\sqrt{3} + \sqrt{3x}}{3 + x}\\)
4. \\(\frac{9 + \sqrt{3x}}{3 - x}\\) – wait, no, the first (leftmost) options fraction is \\(\frac{9 + \sqrt{3x}}{3 - x}\\), then next (to the right) is \\(\frac{\sqrt{3} + \sqrt{3x}}{3 + x}\\), then \\(\frac{9 + \sqrt{3x}}{3 + x}\\), then \\(\frac{3 + \sqrt{3x}}{3 - x}\\) (rightmost). wait, the ocr for the options (as per the images text in the boxes):

first box (leftmost): \\(\frac{9 + \sqrt{3x}}{3 - x}\\)
second box: \\(\frac{\sqrt{3} + \sqrt{3x}}{3 + x}\\)
third box: \\(\frac{9 + \sqrt{3x}}{3 + x}\\)
fourth box (rightmost): \\(\frac{3 + \sqrt{3x}}{3 - x}\\)

Explanation:

Step1: Recall Rationalization

To rationalize \(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{x}}\), multiply numerator and denominator by the conjugate of the denominator, \(\sqrt{3}+\sqrt{x}\).

$$ \frac{\sqrt{3}}{\sqrt{3}-\sqrt{x}} \cdot \frac{\sqrt{3}+\sqrt{x}}{\sqrt{3}+\sqrt{x}} $$

Step2: Multiply Numerator

Multiply the numerators: \(\sqrt{3}(\sqrt{3}+\sqrt{x})=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{x}=3 + \sqrt{3x}\) (using \(\sqrt{a}\cdot\sqrt{a}=a\) and \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)).

Step3: Multiply Denominator

Multiply the denominators: \((\sqrt{3}-\sqrt{x})(\sqrt{3}+\sqrt{x})=(\sqrt{3})^2-(\sqrt{x})^2 = 3 - x\) (using difference of squares: \((a - b)(a + b)=a^2 - b^2\)).

Step4: Form the Simplified Fraction

Combine numerator and denominator: \(\frac{3+\sqrt{3x}}{3 - x}\)

Answer:

\(\frac{3+\sqrt{3x}}{3 - x}\) (corresponding to the first option among the given choices, likely the right - most option in the image with \(\frac{3+\sqrt{3x}}{3 - x}\))