$\frac{3}{4}\left(1 + \frac{1}{3}\right)\left(1 + \frac{2}{3}\right)\left(1 - \frac{2}{5}\right)\left(1 + \frac{6}{7}\right)\left(1 - \frac{12}{13}\right) = x$

### Concept & Formula To solve an expression involving a chain product of fractions, first evaluate the expression inside each set of parentheses, then write out the continuous product and cancel common factors across numerators and denominators. ### Step-by-Step Solution Simplify every individual bracketed term first: $$1 + \frac{1}{3} = \frac{4}{3}$$ $$1 + \frac{2}{3} = \frac{5}{3}$$ $$1 - \frac{2}{5} = \frac{3}{5}$$ $$1 + \frac{6}{7} = \frac{13}{7}$$ $$1 - \frac{12}{13} = \frac{1}{13}$$ Now, substitute these values back into the primary expression: $$\frac{3}{4} \times \frac{4}{3} \times \frac{5}{3} \times \frac{3}{5} \times \frac{13}{7} \times \frac{1}{13}$$ Group and cancel terms sequentially: * $\frac{3}{4} \times \frac{4}{3} = 1$ * $\frac{5}{3} \times \frac{3}{5} = 1$ * $\frac{13}{7} \times \frac{1}{13} = \frac{1}{7}$ Combining all of these results gives: $$1 \times 1 \times \frac{1}{7} = \frac{1}{7}$$ ### Exam Strategy & Shortcut Scan the entire expression for reciprocal pairs. We can clearly see that $\frac{3}{4}$ and $\left(1+\frac{1}{3}\right) = \frac{4}{3}$ are reciprocals, so they cancel out completely. Similarly, $\left(1+\frac{2}{3}\right) = \frac{5}{3}$ and $\left(1-\frac{2}{5}\right) = \frac{3}{5}$ are reciprocals and cancel out. We are immediately left with just the final two components: $\frac{13}{7} \times \frac{1}{13} = \frac{1}{7}$. ### Common Pitfall Avoid trying to find a common denominator for the whole expression or performing cross-multiplication too early. Simplify each component cleanly first to let the cancellations happen naturally. ### Final Answer **Therefore, the correct answer is 1/7.**