已知 $|x-2 y+1|+(x-y-5)^{2}=0$, 据此可列出关于 $x, y$ 的二元一次方程组:
解: $\because|x-2 y+1| \geq 0, \quad(x-y-5)^{2} \geq 0$,

且 $|x-2 y+1|+(x-y-5)^{2}=0$

$\therefore x-2 y+1=0, x-y-5=0$
$\therefore$ 方程组为: $\left\{\begin{array}{c}x-2 y+1=0 \\ x-y-5=0\end{array}\right.$.