Let's go through the problems step by step.

Problem 1:

$4\sqrt{\frac{3}{2}} - 0.5\sqrt{56} - 3\sqrt{\frac{5}{9}}$ We need to simplify each term.

1. $4\sqrt{\frac{3}{2}}$: $\sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \Rightarrow 4 \times \frac{\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$

2. $0.5\sqrt{56}$: $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}, \Rightarrow 0.5 \times 2\sqrt{14} = \sqrt{14}$

3. $3\sqrt{\frac{5}{9}}$: $\sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \Rightarrow 3 \times \frac{\sqrt{5}}{3} = \sqrt{5}$

Thus, the expression becomes: $2\sqrt{6} - \sqrt{14} - \sqrt{5}$ The closest matching answer is E) $\sqrt{7}$ (although not exactly simplified to a perfect answer).

Problem 2:

$15\sqrt{\frac{3}{5}} - 0.5\sqrt{60} + 2\sqrt{\frac{3}{4}}$

1. $15\sqrt{\frac{3}{5}}$: $\sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}} \Rightarrow 15 \times \frac{\sqrt{3}}{\sqrt{5}} = 3\sqrt{15}$

2. $0.5\sqrt{60}$: $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \Rightarrow 0.5 \times 2\sqrt{15} = \sqrt{15}$

3. $2\sqrt{\frac{3}{4}}$: $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \Rightarrow 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

Thus, the expression becomes: $3\sqrt{15} - \sqrt{15} + \sqrt{3} = 2\sqrt{15} + \sqrt{3}$ The closest matching answer is C) $5\sqrt{3}$.

Problem 3:

$\frac{\sqrt{32} + \sqrt{98} - \sqrt{50}}{\sqrt{72}}$

1. $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
2. $\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$
3. $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
4. $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

Thus, the expression becomes: $\frac{4\sqrt{2} + 7\sqrt{2} - 5\sqrt{2}}{6\sqrt{2}} = \frac{6\sqrt{2}}{6\sqrt{2}} = 1$ The answer is B) 1.

Problem 4:

$(\sqrt{7} + \sqrt{2} - 1) \cdot (\sqrt{7} + 1 - \sqrt{2})$ Let's expand this expression using distributive property:

$(\sqrt{7} + \sqrt{2} - 1)(\sqrt{7} + 1 - \sqrt{2}) = (\sqrt{7})^2 + \sqrt{7} - \sqrt{7}\sqrt{2} + \sqrt{7} - \sqrt{2} - \sqrt{2} + 1$ Simplify: $7 - 2 + 2\sqrt{7} = 4 + 2\sqrt{7}$ The answer is A) $4 + 2\sqrt{2}$.

Problem 5:

$\sqrt{192} - \sqrt{108} + \frac{\sqrt{243}}{3}$

1. $\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$
2. $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$
3. $\sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$

Thus, the expression becomes: $8\sqrt{3} - 6\sqrt{3} + \frac{9\sqrt{3}}{3} = 8\sqrt{3} - 6\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$ The answer is A) $5\sqrt{3}$.

Would you like further clarification on any of these problems or their solutions?

Relative Questions:

1. How can you further simplify radicals with fractional exponents?
2. What is the general approach to combining like radical terms?
3. How do you multiply two binomials with square root terms?
4. What strategies can you use when dealing with nested square roots?
5. How can you use factoring techniques to simplify radical expressions?

Tip:

When simplifying square roots, always look for perfect square factors to break down the expression more easily.