1. **State the problem:** Solve the equation $$\left(\frac{7}{13} - \frac{1}{3}\right) - \frac{1}{5} = \left(\frac{105}{\Box} - \frac{65}{\Box}\right) \frac{39}{\Box}$$ where the boxes represent unknown denominators to be found. 2. **Simplify the left-hand side:** Calculate $$\frac{7}{13} - \frac{1}{3}$$ first. Find a common denominator for 13 and 3, which is 39. $$\frac{7}{13} = \frac{7 \times 3}{13 \times 3} = \frac{21}{39}$$ $$\frac{1}{3} = \frac{1 \times 13}{3 \times 13} = \frac{13}{39}$$ Subtract: $$\frac{21}{39} - \frac{13}{39} = \frac{8}{39}$$ 3. **Subtract $$\frac{1}{5}$$ from $$\frac{8}{39}$$:** Find common denominator for 39 and 5, which is 195. $$\frac{8}{39} = \frac{8 \times 5}{39 \times 5} = \frac{40}{195}$$ $$\frac{1}{5} = \frac{1 \times 39}{5 \times 39} = \frac{39}{195}$$ Subtract: $$\frac{40}{195} - \frac{39}{195} = \frac{1}{195}$$ So the left-hand side simplifies to $$\frac{1}{195}$$. 4. **Analyze the right-hand side:** The right side is $$\left(\frac{105}{\Box} - \frac{65}{\Box}\right) \frac{39}{\Box}$$. Since the denominators are the same in the subtraction, let the denominator be $$d$$. Then: $$\frac{105}{d} - \frac{65}{d} = \frac{105 - 65}{d} = \frac{40}{d}$$ Multiply by $$\frac{39}{d}$$: $$\frac{40}{d} \times \frac{39}{d} = \frac{40 \times 39}{d^2} = \frac{1560}{d^2}$$ 5. **Set the left and right sides equal:** $$\frac{1}{195} = \frac{1560}{d^2}$$ Cross-multiply: $$d^2 = 1560 \times 195$$ Calculate: $$1560 \times 195 = 304200$$ So: $$d^2 = 304200$$ 6. **Find $$d$$:** $$d = \sqrt{304200}$$ Factor 304200 to simplify: $$304200 = 100 \times 3042$$ $$\sqrt{304200} = \sqrt{100} \times \sqrt{3042} = 10 \times \sqrt{3042}$$ Since 3042 is not a perfect square, the denominator $$d$$ is $$10 \sqrt{3042}$$. 7. **Final answer:** The denominators in the right-hand side fractions are all $$10 \sqrt{3042}$$. Thus: $$\boxed{d = 10 \sqrt{3042}}$$