The value of $$ \frac{(2.697 - 0.498)^2 + (2.697 + 0.498)^2}{2.697 \times 2.697 + 0.498 \times 0.498} $$ is

### Concept & Formula This problem is an application of a fundamental algebraic identity involving the sum of the squares of two binomials. The governing identity is: $$ (a - b)^2 + (a + b)^2 = 2(a^2 + b^2) $$ ### Step-by-Step Solution Let $a = 2.697$ and $b = 0.498$. By substituting these variables into the given expression, the structure becomes much clearer: $$ \frac{(a - b)^2 + (a + b)^2}{a^2 + b^2} $$ Now, apply the algebraic identity to the numerator: $$ \frac{2(a^2 + b^2)}{a^2 + b^2} $$ Since $(a^2 + b^2)$ is common to both the numerator and the denominator, they cancel each other out completely. The expression simplifies to exactly $2$. ### Exam Strategy & Shortcut Whenever an expression takes the form of the sum of squared binomials over the sum of their squares, do not waste time plugging in numbers. Recognizing the $2(a^2+b^2)$ pattern allows you to bypass arithmetic entirely. The answer will always be $2$, irrespective of what the actual decimal values are. ### Common Pitfall The most common trap is attempting brute-force calculation. Squaring three-digit decimals by hand will drain your exam time and vastly increase the likelihood of a silly arithmetic mistake. Always step back and look for the hidden algebraic framework first. ### Final Answer **Therefore, the correct answer is 2.**