Question

what is \\(\frac{\sqrt{3}}{\sqrt{3} - \sqrt{x}}\\) in simplest form? \\(\frac{3 + \sqrt{3x}}{3 - x}\\) \\(\frac{9 + \sqrt{3x}}{3 - x}\\) \\(\frac{\sqrt{3} + \sqrt{x}}{3 + x}\\)

Explanation:

Step1: Rationalize the denominator

Multiply numerator and denominator by the conjugate $\sqrt{3}+\sqrt{x}$:
$$\frac{\sqrt{3}}{\sqrt{3}-\sqrt{x}} \times \frac{\sqrt{3}+\sqrt{x}}{\sqrt{3}+\sqrt{x}}$$

Step2: Expand numerator

Use distributive property:
$$\sqrt{3}(\sqrt{3}+\sqrt{x}) = (\sqrt{3})^2 + \sqrt{3}\sqrt{x} = 3 + \sqrt{3x}$$

Step3: Expand denominator

Use difference of squares:
$$(\sqrt{3}-\sqrt{x})(\sqrt{3}+\sqrt{x}) = (\sqrt{3})^2 - (\sqrt{x})^2 = 3 - x$$

Step4: Combine numerator and denominator

Form the simplified fraction:
$$\frac{3 + \sqrt{3x}}{3 - x}$$

Answer:

$\boldsymbol{\frac{3 + \sqrt{3x}}{3 - x}}$ (the first option)