$4 \cdot 21$ If $M=3 x^{2}-8 x y+9 y^{2}-4 x+6 y+13$, then the following must be true:
(A) $M \geqslant 0$.
(B) $M0$.
(D) $M \leqslant 0$. (Shanxi Province, China Junior High School Mathematics Competition, 1990)
[Solution]
$$
\begin{aligned}
M & =3 x^{2}-8 x y+9 y^{2}-4 x+6 y+13 \\
& =\left(2 x^{2}-8 x y+8 y^{2}\right)+\left(x^{2}-4 x+4\right)+\left(y^{2}+6 y+9\right) \\
& =2(x-2 y)^{2}+(x-2)^{2}+(y+3)^{2},
\end{aligned}
$$

Since $x=2$ and $y=-3$ cannot both hold at the same time with $x-2 y=0$,
then $M \neq 0$, which means $M>0$.
Therefore, the answer is $(C)$.