Let's break down the given expression step by step:

The expression is:

$\frac{\frac{3}{2^{1/2}} + \frac{5^{1/2}}{3^{1/2}}}{1 + \frac{3}{6^{1/2}}}$

Step 1: Simplify each part.

1. $2^{1/2}$ is the square root of 2, so $2^{1/2} = \sqrt{2}$.
2. $5^{1/2}$ is the square root of 5, so $5^{1/2} = \sqrt{5}$.
3. $3^{1/2}$ is the square root of 3, so $3^{1/2} = \sqrt{3}$.
4. $6^{1/2}$ is the square root of 6, so $6^{1/2} = \sqrt{6}$.

Step 2: Plug in the simplified square roots.

Now the expression becomes:

$\frac{\frac{3}{\sqrt{2}} + \frac{\sqrt{5}}{\sqrt{3}}}{1 + \frac{3}{\sqrt{6}}}$

Step 3: Handle the numerator.

• $\frac{3}{\sqrt{2}}$ is just $3 / \sqrt{2}$.
• $\frac{\sqrt{5}}{\sqrt{3}}$ simplifies to $\sqrt{\frac{5}{3}}$.

Thus, the numerator is:

$\frac{3}{\sqrt{2}} + \sqrt{\frac{5}{3}}$

Step 4: Handle the denominator.

The denominator is:

$1 + \frac{3}{\sqrt{6}}$

Step 5: Simplify the overall expression.

We now have:

$\frac{\frac{3}{\sqrt{2}} + \sqrt{\frac{5}{3}}}{1 + \frac{3}{\sqrt{6}}}$

At this point, we could rationalize the terms, but to keep it simple, let's calculate each part approximately:

• $\frac{3}{\sqrt{2}} \approx 2.121$
• $\sqrt{\frac{5}{3}} \approx 1.291$
• $\frac{3}{\sqrt{6}} \approx 1.225$

Substituting these approximations into the expression:

$\frac{2.121 + 1.291}{1 + 1.225} = \frac{3.412}{2.225} \approx 1.534$

Final Answer:

The approximate value of the expression is 1.534.

Would you like to go over the steps in more detail or have any other questions?

Additional questions:

1. How do you rationalize a denominator like $\frac{3}{\sqrt{2}}$?
2. Can this problem be solved using algebraic identities to simplify square roots?
3. How do approximations affect the final answer in complex expressions like this?
4. What is the importance of simplifying square roots in algebraic expressions?
5. What other methods could be used to solve this expression more precisely?

Tip:

When dealing with square roots in a fraction, consider rationalizing the denominator (multiplying both the numerator and denominator by the square root) to simplify the expression!