1. The problem is to simplify the expressions $\frac{\sqrt{2907}}{51}$ and $\frac{\sqrt{1728}}{18}$.\n\n2. Recall the property of square roots and division: $$\frac{\sqrt{a}}{b} = \sqrt{\frac{a}{b^2}}.$$ This allows us to simplify the expressions by rewriting the denominator inside the square root.\n\n3. For the first expression: $$\frac{\sqrt{2907}}{51} = \sqrt{\frac{2907}{51^2}} = \sqrt{\frac{2907}{2601}}.$$\nCalculate the fraction inside the root: $$\frac{2907}{2601} \approx 1.117.$$\nSince this is not a perfect square, the simplified form is $$\sqrt{\frac{2907}{2601}}.$$\n\n4. For the second expression: $$\frac{\sqrt{1728}}{18} = \sqrt{\frac{1728}{18^2}} = \sqrt{\frac{1728}{324}}.$$\nCalculate the fraction inside the root: $$\frac{1728}{324} = \frac{1728}{324} = 5.333... = \frac{16}{3}.$$\nSo the expression becomes $$\sqrt{\frac{16}{3}} = \frac{\sqrt{16}}{\sqrt{3}} = \frac{4}{\sqrt{3}}.$$\n\n5. Rationalize the denominator: $$\frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}.$$\n\nFinal answers:\n- $$\frac{\sqrt{2907}}{51} = \sqrt{\frac{2907}{2601}}$$ (approx. 1.057)\n- $$\frac{\sqrt{1728}}{18} = \frac{4\sqrt{3}}{3}$$