The problem is to simplify the given expression:

$\frac{4}{7}\sqrt{147} + \frac{3}{8}\sqrt{192} - \frac{1}{5}\sqrt{75}.$

Step 1: Simplify the square root terms

1. $\sqrt{147} = \sqrt{49 \cdot 3} = 7\sqrt{3}$, so $\frac{4}{7}\sqrt{147} = \frac{4}{7}(7\sqrt{3}) = 4\sqrt{3}$.

2. $\sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3}$, so $\frac{3}{8}\sqrt{192} = \frac{3}{8}(8\sqrt{3}) = 3\sqrt{3}$.

3. $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$, so $\frac{1}{5}\sqrt{75} = \frac{1}{5}(5\sqrt{3}) = \sqrt{3}$.

Step 2: Add/Subtract the terms

Now substitute the simplified terms: $4\sqrt{3} + 3\sqrt{3} - \sqrt{3}.$

Combine like terms: $(4 + 3 - 1)\sqrt{3} = 6\sqrt{3}.$

Final Answer:

The simplified result is: $\boxed{6\sqrt{3}}$

The correct option is (4).

Would you like further explanation, or do you have questions?

5 Related Questions:

1. How do you simplify square root expressions involving products?
2. Why is $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ valid for positive numbers $a$ and $b$?
3. Can this method be extended to solve similar problems involving different roots?
4. How can you combine like terms in algebra involving radicals?
5. What are some tips for quickly factoring numbers under square roots?

Tip:

Always check for perfect squares when simplifying square roots—they help to simplify problems significantly!