解下列方程组(1)x2/4-y2=1,x-y-2=0:(2)(x-3)2+y2=9,x+2y=0
解下列方程组:(1)x+13=2y2(x+1)−y=11
解方程组:y2=x3-3x2+2x;x2=y3-3y2+2y
解方程组(1):xy-y-2x+2=0 (2):x2-2x+y2-4x=-5
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x
已知圆x2+y2-2x-2y+1=0求x2+y2的最大值
二元二次方程组x2+y2+2xy=9 (x-y)2-3x+3y+2=0
已知2x+y=7,x2+y2=5,求(4x+2y)2-3x2-y2+2(1-y2)的值.
已知x2+y2+4x-6y+13=0 求x2-2x\x2+3y2
已知x2-4x-1=0,求代数式(2x-3)2-(x+y)(x-y)-y2的值
已知实数x.y满足(x2+y2)(x2+y2-1)=2,求x2+y2的值
已知x2+4x+y2-2y+5=0,则x2+y2=______.
已知方程x2 +y2+4x-2y-4=0,求x2 +y2的最大值