9 -12 Let $m>n>1, \quad a=(\lg m n)^{\frac{1}{2}}-(\lg m)^{\frac{1}{2}}, \quad b=(\lg m)^{\frac{1}{2}}$ $-\left(\lg \frac{m}{n}\right)^{\frac{1}{2}}$. Then the relationship between $a$ and $b$ is
(A) $a>b$.
(B) $a=b$.
(C) $a<b$.
(D) Uncertain
(China Sichuan Province Junior High School Mathematics Competition, 1990)
[Solution] From the given, we have
$$
\frac{1}{a}=\frac{(\lg m n)^{\frac{1}{2}}+(\lg m)^{\frac{1}{2}}}{\lg m n-\lg m}=\frac{(\lg m+\lg n)^{\frac{1}{2}}+(\lg m)^{\frac{1}{2}}}{\lg n} \text {, }
$$

Similarly, $\frac{1}{b}=\frac{(\lg m-\lg n)^{\frac{1}{2}}+(\lg m)^{\frac{1}{2}}}{\lg n}$, and given $\frac{1}{a}>\frac{1}{b}$, and $a>0, b>0$, then $a<b$. Therefore, the answer is $(C)$.