1. We are given $x = 2\sqrt{3}$ and asked to find and rationalise the denominator for the expressions: a) $\frac{x+1}{x}$ b) $\frac{x-1}{x}$ c) $\left(\frac{x+1}{x}\right)^2$ 2. First, substitute $x = 2\sqrt{3}$ into each expression. For a): $$\frac{x+1}{x} = \frac{2\sqrt{3} + 1}{2\sqrt{3}}$$ 3. To rationalise the denominator, multiply numerator and denominator by the conjugate or the denominator's radical part. Here, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $$\frac{2\sqrt{3} + 1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{(2\sqrt{3} + 1)\sqrt{3}}{2\sqrt{3} \times \sqrt{3}}$$ 4. Simplify numerator and denominator: $$\text{Numerator} = 2\sqrt{3} \times \sqrt{3} + 1 \times \sqrt{3} = 2 \times 3 + \sqrt{3} = 6 + \sqrt{3}$$ $$\text{Denominator} = 2 \times 3 = 6$$ So, $$\frac{6 + \sqrt{3}}{6}$$ 5. For b): $$\frac{x-1}{x} = \frac{2\sqrt{3} - 1}{2\sqrt{3}}$$ Multiply numerator and denominator by $\frac{\sqrt{3}}{\sqrt{3}}$: $$\frac{2\sqrt{3} - 1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{(2\sqrt{3} - 1)\sqrt{3}}{6}$$ Simplify numerator: $$2\sqrt{3} \times \sqrt{3} - 1 \times \sqrt{3} = 6 - \sqrt{3}$$ So, $$\frac{6 - \sqrt{3}}{6}$$ 6. For c): $$\left(\frac{x+1}{x}\right)^2 = \left(\frac{2\sqrt{3} + 1}{2\sqrt{3}}\right)^2$$ From step 4, we have: $$\frac{6 + \sqrt{3}}{6}$$ Square this: $$\left(\frac{6 + \sqrt{3}}{6}\right)^2 = \frac{(6 + \sqrt{3})^2}{36}$$ 7. Expand numerator: $$(6 + \sqrt{3})^2 = 6^2 + 2 \times 6 \times \sqrt{3} + (\sqrt{3})^2 = 36 + 12\sqrt{3} + 3 = 39 + 12\sqrt{3}$$ So, $$\frac{39 + 12\sqrt{3}}{36}$$ 8. Simplify fraction by dividing numerator and denominator by 3: $$\frac{\cancel{3}(13 + 4\sqrt{3})}{\cancel{3}12} = \frac{13 + 4\sqrt{3}}{12}$$ **Final answers:** a) $\frac{6 + \sqrt{3}}{6}$ b) $\frac{6 - \sqrt{3}}{6}$ c) $\frac{13 + 4\sqrt{3}}{12}$