2. Prove that \( \frac{2 \sqrt{a}}{a}+\frac{2 b}{\sqrt{a}}=\frac{2 \sqrt{a}(1+b)}{a} \) without the use of a calculator.

To prove that \[ \frac{2 \sqrt{a}}{a} + \frac{2b}{\sqrt{a}} = \frac{2 \sqrt{a}(1 + b)}{a}, \] follow these steps: 1. **Simplify Each Term Separately:** - **First Term:** \[ \frac{2 \sqrt{a}}{a} \] Notice that \( a = (\sqrt{a})^2 \), so: \[ \frac{2 \sqrt{a}}{a} = \frac{2 \sqrt{a}}{(\sqrt{a})^2} = \frac{2}{\sqrt{a}} \] - **Second Term:** \[ \frac{2b}{\sqrt{a}} \] This term is already simplified. 2. **Combine the Simplified Terms:** \[ \frac{2}{\sqrt{a}} + \frac{2b}{\sqrt{a}} = \frac{2(1 + b)}{\sqrt{a}} \] 3. **Express the Combined Term with a Common Denominator:** \[ \frac{2(1 + b)}{\sqrt{a}} = \frac{2 \sqrt{a}(1 + b)}{a} \] Here's why: - Multiply numerator and denominator by \( \sqrt{a} \): \[ \frac{2(1 + b)}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} = \frac{2 \sqrt{a}(1 + b)}{a} \] - Since \( a = (\sqrt{a})^2 \), the denominator simplifies to \( a \). 4. **Final Conclusion:** \[ \frac{2 \sqrt{a}}{a} + \frac{2b}{\sqrt{a}} = \frac{2 \sqrt{a}(1 + b)}{a} \] Hence, the equality is proven. **Summary of Steps:** \[ \frac{2 \sqrt{a}}{a} + \frac{2b}{\sqrt{a}} = \frac{2}{\sqrt{a}} + \frac{2b}{\sqrt{a}} = \frac{2(1 + b)}{\sqrt{a}} = \frac{2 \sqrt{a}(1 + b)}{a} \] Answer: By rewriting 2√a⁄a as 2⁄√a and combining terms, we have  2√a/a + 2b/√a = 2(1 + b)/√a = 2√a(1 + b)/a. Thus, the equality holds.