1. If the equation about $x$
$$
x^{2}+(1-2 \sqrt{2}) x+3 m-\sqrt{2}=0
$$
( $m$ is a rational number) has a rational root, then its other root is $(\quad)$.
(A) $\frac{3}{2}+2 \sqrt{2}$
(B) $\frac{3}{2}-2 \sqrt{2}$
(C) $-\frac{1}{2}+2 \sqrt{2}$
(D) $-\frac{1}{2}-2 \sqrt{2}$
- 1. C.

Transform the original equation into
$$
x^{2}+x+3 m=\sqrt{2}(2 x+1) \text {. }
$$

Let one rational root of the original equation be $a$. Then
$$
a^{2}+a+3 m=\sqrt{2}(2 a+1) \text {. }
$$

Since the left side of the above equation is a rational number, the right side must also be a rational number. Therefore,
$$
\left\{\begin{array} { l } 
{ 2 a + 1 = 0 , } \\
{ a ^ { 2 } + a + 3 m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-\frac{1}{2}, \\
m=\frac{1}{12} .
\end{array}\right.\right.
$$

Let the other root of the original equation be $b$. By Vieta's formulas, we get
$$
-\frac{1}{2}+b=-(1-2 \sqrt{2}) \Rightarrow b=-\frac{1}{2}+2 \sqrt{2} .
$$