a) $$ \frac{\alpha^{2}-2 \alpha}{4 a^{3}-8 a^{2}} $$ ק) $$ \frac{x^{2} y+2 y^{2}}{x^{3}+2 x y} $$ ү) $$ \frac{2 \beta^{2}+6 \alpha \beta}{12 \alpha^{2} \beta^{2}+4 \alpha \beta^{3}} $$ б) $$ \frac{\alpha^{2}-4 \beta^{2}}{\alpha^{2}-2 \alpha \beta} $$

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Certainly! Let's simplify each of the given algebraic expressions step by step.

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### **a) $$ \frac{\alpha^{2} - 2 \alpha}{4 a^{3} - 8 a^{2}} $$**

**Step 1: Factor both numerator and denominator.**

- **Numerator:**  
  $$ \alpha^{2} - 2\alpha = \alpha(\alpha - 2) $$

- **Denominator:**  
  $$ 4a^{3} - 8a^{2} = 4a^{2}(a - 2) $$

**Step 2: Write the expression with factored terms.**

$$
\frac{\alpha(\alpha - 2)}{4a^{2}(a - 2)}
$$

**Step 3: Simplify by canceling common factors.**

- If **$$\alpha$$** and **$$a$$** are **independent variables**, there are no common factors to cancel.  
  **Simplified Form:**  
  $$
  \frac{\alpha(\alpha - 2)}{4a^{2}(a - 2)}
  $$

- If **$$\alpha = a$$**, then:
  $$
  \frac{a(a - 2)}{4a^{2}(a - 2)} = \frac{1}{4a}
  $$
  (Assuming $$a \neq 2$$ and $$a \neq 0$$)

---

### **ק) $$ \frac{x^{2} y + 2 y^{2}}{x^{3} + 2 x y} $$**

**Step 1: Factor both numerator and denominator.**

- **Numerator:**  
  $$ x^{2} y + 2 y^{2} = y(x^{2} + 2y) $$

- **Denominator:**  
  $$ x^{3} + 2x y = x(x^{2} + 2y) $$

**Step 2: Write the expression with factored terms.**

$$
\frac{y(x^{2} + 2y)}{x(x^{2} + 2y)}
$$

**Step 3: Simplify by canceling the common factor $$(x^{2} + 2y)$$.**

$$
\frac{y}{x} \quad \text{(provided that } x^{2} + 2y \neq 0\text{)}
$$

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### **ү) $$ \frac{2 \beta^{2} + 6 \alpha \beta}{12 \alpha^{2} \beta^{2} + 4 \alpha \beta^{3}} $$**

**Step 1: Factor both numerator and denominator.**

- **Numerator:**  
  $$ 2\beta^{2} + 6\alpha\beta = 2\beta(\beta + 3\alpha) $$

- **Denominator:**  
  $$ 12\alpha^{2}\beta^{2} + 4\alpha\beta^{3} = 4\alpha\beta^{2}(3\alpha + \beta) $$

**Step 2: Write the expression with factored terms.**

$$
\frac{2\beta(\beta + 3\alpha)}{4\alpha\beta^{2}(3\alpha + \beta)}
$$

**Step 3: Simplify by canceling common factors.**

- Note that $$(\beta + 3\alpha) = (3\alpha + \beta)$$

$$
\frac{2\beta}{4\alpha\beta^{2}} = \frac{2}{4\alpha\beta} = \frac{1}{2\alpha\beta}
$$

**Simplified Form:**  
$$
\frac{1}{2\alpha\beta}
$$

---

### **б) $$ \frac{\alpha^{2} - 4 \beta^{2}}{\alpha^{2} - 2 \alpha \beta} $$**

**Step 1: Factor both numerator and denominator.**

- **Numerator:**  
  $$ \alpha^{2} - 4\beta^{2} = (\alpha - 2\beta)(\alpha + 2\beta) $$

- **Denominator:**  
  $$ \alpha^{2} - 2\alpha\beta = \alpha(\alpha - 2\beta) $$

**Step 2: Write the expression with factored terms.**

$$
\frac{(\alpha - 2\beta)(\alpha + 2\beta)}{\alpha(\alpha - 2\beta)}
$$

**Step 3: Simplify by canceling the common factor $$(\alpha - 2\beta)$$.**

$$
\frac{\alpha + 2\beta}{\alpha} = 1 + \frac{2\beta}{\alpha} \quad \text{(provided that } \alpha \neq 0 \text{ and } \alpha \neq 2\beta\text{)}
$$

**Simplified Form:**  
$$
1 + \frac{2\beta}{\alpha} \quad \text{or} \quad \frac{\alpha + 2\beta}{\alpha}
$$

---

**Summary of Simplified Expressions:**

1. **a)**  
   $$
   \frac{\alpha^{2} - 2 \alpha}{4 a^{3} - 8 a^{2}} = \frac{\alpha(\alpha - 2)}{4a^{2}(a - 2)} \quad \text{(if } \alpha \neq 2 \text{ and } a \neq 2\text{)}
   $$
   - If $$\alpha = a$$:
     $$
     \frac{1}{4a}
     $$

2. **ק)**  
   $$
   \frac{x^{2} y + 2 y^{2}}{x^{3} + 2 x y} = \frac{y}{x} \quad \text{(provided } x^{2} + 2y \neq 0\text{)}
   $$

3. **ү)**  
   $$
   \frac{2 \beta^{2} + 6 \alpha \beta}{12 \alpha^{2} \beta^{2} + 4 \alpha \beta^{3}} = \frac{1}{2\alpha\beta}
   $$

4. **б)**  
   $$
   \frac{\alpha^{2} - 4 \beta^{2}}{\alpha^{2} - 2 \alpha \beta} = 1 + \frac{2\beta}{\alpha} \quad \text{or} \quad \frac{\alpha + 2\beta}{\alpha}
   $$

Feel free to ask if you need further clarification on any of these steps!
Here are the simplified forms of the given expressions: 1. **a)** $$ \frac{\alpha^{2} - 2 \alpha}{4 a^{3} - 8 a^{2}} = \frac{\alpha(\alpha - 2)}{4a^{2}(a - 2)} $$ - If $$\alpha = a$$: $$ \frac{1}{4a} $$ 2. **ק)** $$ \frac{x^{2} y + 2 y^{2}}{x^{3} + 2 x y} = \frac{y}{x} $$ 3. **ү)** $$ \frac{2 \beta^{2} + 6 \alpha \beta}{12 \alpha^{2} \beta^{2} + 4 \alpha \beta^{3}} = \frac{1}{2\alpha\beta} $$ 4. **б)** $$ \frac{\alpha^{2} - 4 \beta^{2}}{\alpha^{2} - 2 \alpha \beta} = 1 + \frac{2\beta}{\alpha} \quad \text{or} \quad \frac{\alpha + 2\beta}{\alpha} $$ These are the simplified forms of the algebraic expressions provided.

a) \( \frac{\alpha^{2}-2 \alpha}{4 a^{3}-8 a^{2}} \) ק) \( \frac{x^{2} y+2 y^{2}}{x^{3}+2 x y} \) ү) \( \frac{2 \beta^{2}+6 \alpha \beta}{12 \alpha^{2} \beta^{2}+4 \alpha \beta^{3}} \) б) \( \frac{\alpha^{2}-4 \beta^{2}}{\alpha^{2}-2 \alpha \beta} \)

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