$\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}}$ is equal to

### Concept & Formula This requires standard **rationalization of a binomial surd denominator**. To rationalize a fraction, multiply both the top and bottom by the conjugate of the denominator. If the denominator is $\sqrt{a} - \sqrt{b}$, the conjugate is $\sqrt{a} + \sqrt{b}$. This leverages the difference of squares: $$ (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b $$ The numerator will form a perfect square binomial: $$ (x + y)^2 = x^2 + 2xy + y^2 $$ ### Step-by-Step Solution * **Calculation / Deduction:** * The given expression is: $$ \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} $$ * Multiply the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{7} + \sqrt{5})$: $$ = \frac{(\sqrt{7} + \sqrt{5})(\sqrt{7} + \sqrt{5})}{(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5})} $$ $$ = \frac{(\sqrt{7} + \sqrt{5})^2}{(\sqrt{7})^2 - (\sqrt{5})^2} $$ * Expand the numerator using the $(x+y)^2$ identity and evaluate the denominator: $$ = \frac{(\sqrt{7})^2 + 2(\sqrt{7})(\sqrt{5}) + (\sqrt{5})^2}{7 - 5} $$ * Simplify the expression: $$ = \frac{7 + 2\sqrt{35} + 5}{2} $$ $$ = \frac{12 + 2\sqrt{35}}{2} $$ * Divide both terms in the numerator by the common denominator $2$: $$ = 6 + \sqrt{35} $$ ### Exam Strategy & Shortcut For any fraction of the specific form $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}$, there is a direct shortcut formula: $$ \text{Result} = \frac{a + b}{a - b} + \frac{2\sqrt{ab}}{a - b} $$ Here, $a=7$ and $b=5$. The difference is $7 - 5 = 2$. The sum is $7 + 5 = 12$. The product is $7 \times 5 = 35$. Substituting these in gives $\frac{12}{2} + \frac{2\sqrt{35}}{2} = 6 + \sqrt{35}$. You can compute this mentally in seconds. ### Common Pitfall Students often incorrectly cancel out terms across the fraction before rationalizing (e.g., trying to cancel $\sqrt{7}$ with $\sqrt{7}$). You cannot cancel parts of a sum or difference in a fraction; you must rationalize first. ### Final Answer **Therefore, the correct answer is $6 + \sqrt{35}$.**