To solve the problem of finding the equivalent fraction for [tex]$ \frac{4}{4-\sqrt{6 x}} $[/tex] by rationalizing the denominator and simplifying, follow these steps:
### Step 1: Rationalize the Denominator
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\( 4 - \sqrt{6x} \)[/tex] is [tex]\( 4 + \sqrt{6x} \)[/tex].
Multiply the numerator and the denominator by [tex]\( 4 + \sqrt{6x} \)[/tex]:
[tex]\[
\frac{4}{4-\sqrt{6 x}} \cdot \frac{4+\sqrt{6 x}}{4+\sqrt{6 x}}
= \frac{4(4+\sqrt{6 x})}{(4-\sqrt{6 x})(4+\sqrt{6 x})}
\][/tex]
### Step 2: Simplify the Numerator
Expand the numerator:
[tex]\[
4(4 + \sqrt{6x})
= 4 \cdot 4 + 4 \cdot \sqrt{6x}
= 16 + 4\sqrt{6x}
\][/tex]
### Step 3: Simplify the Denominator
Use the difference of squares formula for the denominator:
[tex]\[
(4 - \sqrt{6x})(4 + \sqrt{6x})
= 4^2 - (\sqrt{6x})^2
= 16 - 6x
\][/tex]
### Step 4: Combine the Results
Substitute back the simplified numerator and denominator:
[tex]\[
\frac{16 + 4\sqrt{6x}}{16 - 6x}
\][/tex]
### Step 5: Compare with Given Choices
Look at the given choices:
- A. [tex]\( \frac{8 + 2\sqrt{6x}}{16 - 6x} \)[/tex]
- B. [tex]\( \frac{2 + \sqrt{6x}}{4 - 6x} \)[/tex]
- C. [tex]\( \frac{8 + 2\sqrt{6x}}{8 - 3x} \)[/tex]
- D. [tex]\( \frac{8 + 2\sqrt{6x}}{8 - 6x} \)[/tex]
Simplifying and examination show that:
[tex]\[
16 + 4\sqrt{6x} = 8 + 2(8 + 2\sqrt{6x})
\][/tex]
Combining the numerator and the constant:
[tex]\[
\frac{8 + 2\sqrt{6x}}{16 - 6x}
\][/tex]
So the fraction equivalent to [tex]\( \frac{16 + 4 \sqrt{6x}}{16 - 6x} \)[/tex] matches choice A.
Thus, the answer is [tex]\( \boxed{A} \)[/tex].
