1983 AMC 12 Problem 21

Compare the squares to determine which listed differences are positive

Rationalize each positive difference using $a-b=\frac{a^2-b^2}{a+b}$

Since $100\gt99,$ choice A is positive; since $324\lt325,$ choice C is negative; and since $2601\gt2600,$ choice D is positive while E is negative. The two positive values satisfy $$\begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}}. \end{aligned}$$ The second denominator is larger, so choice D is the smaller positive number.

Therefore, the correct answer is D.