8. When $a$ and $b$ are two distinct positive numbers, among the following three algebraic expressions:
甲: $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)$,
乙: $\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}$,
丙: $\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}$

the one with the largest value is
(A) definitely 甲;
(B) definitely 乙;
(C) definitely 丙;
(D) generally not certain, and depends on the values of $a$ and $b$.
8. (D)
\[
\begin{aligned}
\text { A }=\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right) & =a b+\frac{1}{a b}+\frac{b}{a}+\frac{a}{b} \\
& >a b+\frac{1}{a b}+2 . \\
& =\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}=B .
\end{aligned}
\]
\(\therefore\) A \(>\) B. Therefore, the largest cannot be B, eliminating (B). Only need to compare A and C.
Taking special values \(a=1, b=2\) we get
\[
\text { A }=(1+1)\left(2+\frac{1}{2}\right)=5 \text {, C }=\left(\frac{3}{2}+\frac{2}{3}\right)^{2}=4 \frac{25}{36} \text {. }
\]
\(\therefore\) A \(>\) C, therefore, C will not be the largest, eliminating (B).
Let \(a=1, b=5\),
\[
\begin{array}{l}
\text { A }=(1+1)\left(5+\frac{1}{5}\right)=10 \frac{2}{5} . \\
\text { C }=\left(\frac{6}{2}+\frac{2}{6}\right)^{2}=11 \frac{1}{9} .
\end{array}
\]

So in this case A \(<\) C, so A will not take the maximum value, eliminating (A).