QUESTION 2 2.1. Simplify fully, WITHOUT using a calculator: 2.1.1. $$ \frac{12^{n+1} \cdot 27^{n-2}}{18^{2 n-1}} $$ 2.1.2. $$ 2 \sqrt{8}-4 \sqrt{32}+3 \sqrt{50} $$ 2.2. WITHOUT using a calculator, show that $$ \frac{2}{1+\sqrt{2}} $$ 2.3. Solve for $$ x $$, WITHOUT using a calculator: $$ x^{\frac{3}{4}} $$

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Let's solve each part step by step.

### 2.1. Simplify fully

#### 2.1.1. $$ \frac{12^{n+1} \cdot 27^{n-2}}{18^{2 n-1}} $$

First, we will express each base in terms of its prime factors:

- $$ 12 = 2^2 \cdot 3^1 $$
- $$ 27 = 3^3 $$
- $$ 18 = 2^1 \cdot 3^2 $$

Now, substituting these into the expression:

$$
\frac{(2^2 \cdot 3^1)^{n+1} \cdot (3^3)^{n-2}}{(2^1 \cdot 3^2)^{2n-1}}
$$

This simplifies to:

$$
\frac{2^{2(n+1)} \cdot 3^{n+1} \cdot 3^{3(n-2)}}{2^{2n-2} \cdot 3^{2(2n-1)}}
$$

Now, combine the exponents:

$$
= \frac{2^{2n+2} \cdot 3^{n+1 + 3(n-2)}}{2^{2n-2} \cdot 3^{4n-2}}
$$

Calculating the exponent for $$ 3 $$:

$$
n + 1 + 3(n - 2) = n + 1 + 3n - 6 = 4n - 5
$$

So we have:

$$
= \frac{2^{2n+2}}{2^{2n-2}} \cdot \frac{3^{4n-5}}{3^{4n-2}}
$$

Now simplify each part:

$$
= 2^{(2n+2) - (2n-2)} \cdot 3^{(4n-5) - (4n-2)} = 2^{4} \cdot 3^{-3}
$$

Thus, we can write:

$$
= \frac{16}{27}
$$

#### 2.1.2. $$ 2 \sqrt{8}-4 \sqrt{32}+3 \sqrt{50} $$

First, simplify each square root:

- $$ \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} $$
- $$ \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} $$
- $$ \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} $$

Now substitute back into the expression:

$$
2(2\sqrt{2}) - 4(4\sqrt{2}) + 3(5\sqrt{2})
$$

This simplifies to:

$$
4\sqrt{2} - 16\sqrt{2} + 15\sqrt{2}
$$

Combining the terms:

$$
(4 - 16 + 15)\sqrt{2} = 3\sqrt{2}
$$

### 2.2. Show that $$ \frac{2}{1+\sqrt{2}} $$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:

$$
\frac{2}{1+\sqrt{2}} \cdot \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{2(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}
$$

Calculating the denominator:

$$
(1+\sqrt{2})(1-\sqrt{2}) = 1 - 2 = -1
$$

Thus, we have:

$$
= \frac{2(1-\sqrt{2})}{-1} = -2(1-\sqrt{2}) = -2 + 2\sqrt{2}
$$

### 2.3. Solve for $$ x $$, $$ x^{\frac{3}{4}} $$

To solve for $$ x $$, we need to isolate $$ x $$:

$$
x^{\frac{3}{4}} = 1
$$

Raise both sides to the power of $$ \frac{4}{3} $$:

$$
x = 1^{\frac{4}{3}} = 1
$$

### Final Answers

- **2.1.1:** $$ \frac{16}{27} $$
- **2.1.2:** $$ 3\sqrt{2} $$
- **2.2:** $$ -2 + 2\sqrt{2} $$
- **2.3:** $$ x = 1 $$
- **2.1.1:** $$ \frac{16}{27} $$ - **2.1.2:** $$ 3\sqrt{2} $$ - **2.2:** $$ -2 + 2\sqrt{2} $$ - **2.3:** $$ x = 1 $$

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