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(C) Given equations are $x^2-3|x|+2=0$ and $y^2-3y+2=0$. Solving $x^2-3|x|+2=0$: $(|x|-2)(|x|-1)=0$,so $|x|=2$ or $|x|=1$,which gives $x=\pm 2, \pm 1$

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$x+y=3, x-y=-3$

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(C) Given equations are $x^2-3|x|+2=0$ and $y^2-3y+2=0$. Solving $x^2-3|x|+2=0$: $(|x|-2)(|x|-1)=0$,so $|x|=2$ or $|x|=1$,which gives $x=\pm 2, \pm 1$. Solving $y^2-3y+2=0$: $(y-2)(y-1)=0$,so $y=2, 1$. The lines form two squares with vertices: Square $1$: $A(-1, 1), B(-1, 2), C(-2, 2), D(-2, 1)$. Square $2$: $A'(2, 1), B'(2, 2), C'(1, 2), D'(1, 1)$. The diagonals of these squares are lines like $x+y=k$ and $x-y=k$. For the square $ABCD$ formed by these diagonals,the sides are parallel to the axes or at $45^\circ$. Checking the options,the equations $x+y=3$ and $x-y=-3$ represent lines that can form the sides of such a square. Thus,the correct option is $C$.

If the lines drawn along the diagonals of the two squares formed by two pairs of lines $x^2-3|x|+2=0$ and $y^2-3y+2=0$ form a square $ABCD$,then the equations of two adjacent sides of the square $ABCD$ are

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