x2+x+y2-y-2xy=?
x2-(a+b)xy+aby2 x2+4xy-5y2 x2-5x+3 2x2+xy-y2-4x+5y-6
设二元函数z=x2+xy+y2—x-y,x2+y2≤1,求它的最大值和最小值.
由X2-2XY+Y2-X+Y-1=0得到X-Y的值为.
若实数xy满足x2十y2-2x+4y=0,则x-2y的最大值为多少?
已知x3+y3=(x+y)(x2-xy+y2)称为立方和公式,x3-y3=(x-y)(x2+xy+y2)称为立方差公式,
已知xy为实数求x2-2xy+6y2-14x-6y+72的最小值
若x2y+xy2=30,xy=6,求下列代数式的值:(1)x2+y2;(2)x-y.
x2+y2+xy-3y+3=0 求x的y次方
已知x2+y2+2x-6y+10=0,求xy的值
已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值
若|x+2|+|y-3|=0,则x2+y2=________
x2-xy-2y2-x+5y-2