已知4x^2+3y^2-8x+12y+16=0求{x/(x+y)-x^2/(x^2+2xy+y^2)}/{x/(x+y)
1个回答
(2X-2)²+3(y+2)²=0
x=1
y=-2
相关问题
已知x+y=8,xy=12,求(1)x2y+xy2(2)x2-xy+y2(3)x-y
已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+
已知:3x2-xy-4y2=0(x≠y)求(x—2y)/(x+2y)
已知x^2+4y^2-4x+4y+5=0,求(x^4-y^4/2x^2+xy-y^2)*(2x-y/xy-y^2)/(x
已知x^2+y^2-2x+4y+5=0.求{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-
x^2-4xy+4y^2=16(x+y)^2-x-y-12=0
已知(y+2x)/xy=3/2,求(4x+3xy+2y)/(-4x+8xy-2y)
已知(x+3)^2+|x+y+5|=0,求3x^2y+|-2x^2y-[-2xy+(x^2y-4x^2)]-xy|的值
已知:x2-y2=16,x+y=2,求:3x-3y/y2-x2+x2-2xy+y2/x2-y2-x2-x-6/x+2除以
已知x^2+y^2-2x+4y+5=0,求(x^4-y^4)/(x^2-2xy+y^2)*(x-y