9. If $-\sqrt{\frac{1}{(a-b)^{2}}} \cdot(b-a)=1$, then the solution set of the system of linear inequalities in $\mathrm{x}$ $\left\{\begin{array}{l}3 x-4 a \leq a-2 x, \\ \frac{3 x+2 b}{5}>b\end{array}\right.$ is $(\quad)$
(A) $x \leq a$
(B) $x>b$
(C) $b<x \leq a$
(D) No solution
Answer: (C)
Solution: Since $-\sqrt{\frac{1}{(a-b)^{2}}} \cdot(b-a)=1$, it follows that $a>b$.
Solving the inequality $3 x-4 a \leq a-2 x$, we get
$x \leq a$,

Solving the inequality $\frac{3 x+2 b}{5}>b$, we get
$x>b$,

Therefore, the solution set of the system of inequalities is
$b<x \leq a$.

Hence, the correct choice is (C).