Q. Show that \( \frac{\sqrt{x}}{x}+\frac{y}{\sqrt{x}} \) can be written as \( \frac{\sqrt{x}(y+1)}{x} \) where \( x>0 \).

To show that \[ \frac{\sqrt{x}}{x} + \frac{y}{\sqrt{x}} \] can be rewritten as \[ \frac{\sqrt{x}(y + 1)}{x}, \] where \( x > 0 \), follow these steps: 1. **Simplify the First Term:** \[ \frac{\sqrt{x}}{x} = \frac{\sqrt{x}}{(\sqrt{x})^2} = \frac{1}{\sqrt{x}}. \] This is because \( x = (\sqrt{x})^2 \). 2. **Rewrite the Original Expression:** Substitute the simplified first term back into the original expression: \[ \frac{1}{\sqrt{x}} + \frac{y}{\sqrt{x}}. \] 3. **Factor Out the Common Denominator:** Since both terms have the same denominator \( \sqrt{x} \), factor it out: \[ \frac{1 + y}{\sqrt{x}}. \] 4. **Express as a Single Fraction:** Notice that \( \frac{1 + y}{\sqrt{x}} \) can be written as: \[ \frac{\sqrt{x}(1 + y)}{x}. \] This is because \( \sqrt{x} \times \sqrt{x} = x \), so multiplying the numerator and denominator by \( \sqrt{x} \) maintains equality: \[ \frac{1 + y}{\sqrt{x}} = \frac{\sqrt{x}(1 + y)}{\sqrt{x} \times \sqrt{x}} = \frac{\sqrt{x}(1 + y)}{x}. \] 5. **Final Expression:** Thus, the original expression can be written as: \[ \frac{\sqrt{x}(y + 1)}{x}. \] **Conclusion:** \[ \frac{\sqrt{x}}{x} + \frac{y}{\sqrt{x}} = \frac{\sqrt{x}(y + 1)}{x} \]