We are given the expression:

$\frac{\left( (7 - 3x)^{\frac{1}{2}} + \frac{3x}{2} (7 - 3x)^{-\frac{1}{2}} \right)}{7 - 3x}$

Let's simplify it step by step.

Step 1: Factor Out $(7 - 3x)^{-\frac{1}{2}}$

Rewriting the terms:

$(7 - 3x)^{\frac{1}{2}} = (7 - 3x)^{\frac{1}{2}}$

$\frac{3x}{2} (7 - 3x)^{-\frac{1}{2}} = \frac{3x}{2} \cdot \frac{1}{(7 - 3x)^{\frac{1}{2}}}$

So, rewriting the numerator:

$(7 - 3x)^{\frac{1}{2}} + \frac{3x}{2} (7 - 3x)^{-\frac{1}{2}}$

Factoring out $(7 - 3x)^{-\frac{1}{2}}$:

$(7 - 3x)^{-\frac{1}{2}} \left( (7 - 3x) + \frac{3x}{2} \right)$

Step 2: Simplify the Bracketed Term

$(7 - 3x) + \frac{3x}{2} = \frac{2(7 - 3x)}{2} + \frac{3x}{2}$

$= \frac{14 - 6x + 3x}{2} = \frac{14 - 3x}{2}$

So the numerator becomes:

$(7 - 3x)^{-\frac{1}{2}} \cdot \frac{14 - 3x}{2}$

Step 3: Divide by $7 - 3x$

Since we are dividing by $7 - 3x$, rewrite it as:

$\frac{(7 - 3x)^{-\frac{1}{2}} \cdot \frac{14 - 3x}{2}}{7 - 3x}$

Rewriting $7 - 3x$ as $(7 - 3x)^1$:

$(7 - 3x)^{-\frac{1}{2}} \cdot \frac{14 - 3x}{2} \cdot (7 - 3x)^{-1}$

Using exponent properties:

$(7 - 3x)^{-\frac{1}{2} - 1} \cdot \frac{14 - 3x}{2}$

$(7 - 3x)^{-\frac{3}{2}} \cdot \frac{14 - 3x}{2}$

Since $(7 - 3x)^{-\frac{3}{2}}$ can be rewritten as:

$\frac{1}{(7 - 3x)^{\frac{3}{2}}}$

The final expression simplifies to:

$\frac{14 - 3x}{2(7 - 3x)^{\frac{3}{2}}}$

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Would you like me to verify or explain any part in more detail?

Related Questions:

1. How do you simplify expressions with fractional exponents?
2. What are the rules for factoring out common terms in expressions?
3. How do you handle negative exponents when simplifying algebraic expressions?
4. What are some common strategies for simplifying complex fractions?
5. How do exponent rules apply to rational expressions?

Tip: When dealing with negative exponents, rewrite them as fractions to simplify expressions easily.