AGE_NT_HEADERSn=0;line(x 2,y 1,x 68,y 1);

line(x 8,y 11,x 12,y 11);[save_y change_y-1];putimage(x,y,lo 请问：line(x,y 2,x,y 94);dLine,//com 设x>1,y>0,若x^y+x^-y=2根号2,则x^y-x^-y等于 1、x(x-y)(x+y)-x(x+y)^2 若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)& 已知x,y满足约束条件：x-y+1>=0,x+y-2>=0,x 若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值. 先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y (1)(x^2/x)-y-x-y 已知(2x+1)*2+y*2+2y+1=0 求{(x*2+y*2)-(x-y)*2+2y(x-y)}/(2y) (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1