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) \( \frac{\alpha^{2}-4 \beta^{2}}{\alpha^{2}-2 \alpha \beta} \)

Certainly! Let's simplify each of the given algebraic expressions step by step. --- ### **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 \((x^{2} + 2y)\).** \[ \frac{y}{x} \quad \text{(provided that } x^{2} + 2y \neq 0\text{)} \] --- ### **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!