已知x﹙x+1﹚-﹙x2+y﹚=﹣1,则x2+y2/2-xy=﹙ ﹚
设二元函数z=x2+xy+y2—x-y,x2+y2≤1,求它的最大值和最小值.
请问:已知实数x,y满足1≤x2+y2≤4,求u=x2+xy+y2的最大值和最小值
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,
由X2-2XY+Y2-X+Y-1=0得到X-Y的值为.
已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x
已知x-y+1,X2+Y2=25 求(x+y)2和x2-xy+y2的值
x2-(a+b)xy+aby2 x2+4xy-5y2 x2-5x+3 2x2+xy-y2-4x+5y-6
已知x(x-1)+(y-x2)=6,则代数式x2+y2/2-xy的值是
已知两圆x2+y2-2x-6y-1=0和x2+y2-10x-12y+m=0(1)m取何值时两圆外切
若实数xy满足x2十y2-2x+4y=0,则x-2y的最大值为多少?
化简x2+2x+1分之x2-y2÷x+1分之x-y