1. **Problem 1:** Given $x = \sqrt{7} - \sqrt{5}$, express $\frac{1}{x}$ in the form $\sqrt{a} + \sqrt{b}$. Step 1: Start with $\frac{1}{x} = \frac{1}{\sqrt{7} - \sqrt{5}}$. Step 2: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{7} + \sqrt{5}$: $$\frac{1}{\sqrt{7} - \sqrt{5}} \times \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \frac{\sqrt{7} + \sqrt{5}}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{\sqrt{7} + \sqrt{5}}{7 - 5} = \frac{\sqrt{7} + \sqrt{5}}{2}$$ Step 3: So, $\frac{1}{x} = \frac{\sqrt{7} + \sqrt{5}}{2}$. 2. **Problem 2:** Can a non-terminating, non-repeating decimal be expressed as a rational number? Step 1: No, a non-terminating and non-repeating decimal cannot be expressed as a rational number because rational numbers have decimal expansions that either terminate or repeat. Step 2: For example, the decimal expansion of $\pi = 3.141592\ldots$ is non-terminating and non-repeating, and $\pi$ cannot be expressed as a ratio of integers, hence it is irrational. 3. **Problem 3:** Given $\alpha$ and $\beta$ are zeros of $p(x) = 2x^2 - 5x + k$, and $\alpha^2 + \beta^2 = 21$, find $k$. Step 1: Use the relationships from the quadratic: Sum of zeros $\alpha + \beta = \frac{5}{2}$ (since coefficient of $x$ is -5 and leading coefficient is 2). Product of zeros $\alpha \beta = \frac{k}{2}$. Step 2: Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$. Given $\alpha^2 + \beta^2 = 21$, substitute: $$21 = \left(\frac{5}{2}\right)^2 - 2 \times \frac{k}{2} = \frac{25}{4} - k$$ Step 3: Solve for $k$: $$k = \frac{25}{4} - 21 = \frac{25}{4} - \frac{84}{4} = -\frac{59}{4}$$ Step 4: So, $k = -\frac{59}{4}$.