x2次-xy+2x+y-3
x2-2xy+y2+3x-3y+2.
2x2+xy-3y2+x+4y-1因式分解
[(3x的2次幂y+2xy的2次幂)除以xy]的2次幂=
已知x2+y2+2x-8y+17=0,求x的2005次+xy的值
先化简再求值2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)],其中x=-1,y=-2.
x(x-1)-(x2-y)=-3,求x2-y2-2xy的值
x(x-1)-(x2-y)=-3,求x2+y2-2xy的值
已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值
因式分解x2+y-xy-x
因式分解方法x2-xy+2x-y-3是x方-xy+2x-y-3,
已知x+y=4,xy=2,则x2+y2+3xy=______.
已知x+y=4,xy=2,求x2+y2+3xy的值.