1.59 Given the simplest radical form $a \sqrt{2 a+b}$ and $\sqrt[a-b]{7}$ are like radicals, then the values of $a, b$ that satisfy the condition are
(A) do not exist.
(B) there is one set.
(C) there are two sets.
(D) there are more than two sets.
(China Junior High School Mathematics League, 1989)
[Solution] From the given conditions and the properties of similar radicals, we have 
$$
\left\{\begin{array}{l}
2 a+b=7, \\
a-b=2 \text{. }
\end{array}\right.
$$
Solving this, we get the unique solution:
$$
\left\{\begin{array}{l}
a=3, \\
b=1 .
\end{array}\right.
$$

Therefore, the correct choice is $(B)$.