a) $\dfrac{99 × 100 + 98}{100 × 101 - 102}$
= $\dfrac{99 × 100 + (99 - 1)}{100 × 101 - (101 + 1}$
= $\dfrac{99 × 101 - 1}{99 × 101 - 1}$
= $\dfrac{99 - 1}{99 - 1}$
= 1
b) $\dfrac{327 × 329 - 1}{396 + 327 × 328}$
= $\dfrac{327 × (328 + 1) - 1}{396 + 327 × 328}$
= $\dfrac{327 × 328 + (327 - 1)}{396 + 327 × 328}$
= $\dfrac{327 × 328 + 326}{396 + 327 × 328}$
= 1