1. Given the quadratic function $f(x)=x^{2}-3 x+2$. Then the number of distinct real roots of the equation $f(f(x))=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
-1.D.
Since $f(f(x))=x^{4}-6 x^{3}+10 x^{2}-3 x$, then $x(x-3)\left(x^{2}-3 x+1\right)=0$.
Solving, we get $x_{1}=0, x_{2}=3, x_{3,4}=\frac{3 \pm \sqrt{5}}{2}$.
Therefore, the original equation has 4 distinct real roots.