x=2 y=4
|x-y+2|+(y-4)^2=0,求 1/2(x-y))^3-4xy-x^5=?
2个回答
相关问题
-
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+y=3,xy=-2,求:1.(3x-4y+2xy)-(2x-5y+5xy)的值;2.(3x-5y+4xy)-(
-
(x-1)²+ │y+1│=0 求2x²y-(2x²y-4x²y+5xy
-
若x^4+y^4=15,xy^2-x^2y=-5,求x^4-y^4+3xy^2-x^2y-2xy^2+2y^4
-
①﹙2x+5﹚/﹙3x-6﹚+﹙4-5x﹚/﹙2x-4﹚=0②若2x²-3xy+y²=0求x/y+y
-
x^2-2xy+y^2=14x^2-4xy+y^2=46x^2-5xy+y^2=0x^2-4xy+3y^2+4x-8y+
-
(-x^2+3xy-0.5y^2)-(0.5x^2+4xy-1.5y^2)=-0.5x^2+( )+y^2
-
已知(x+2)²+/y+1/=0求5xy²-2x²y+【3xy²-(4xy²-2x²y)】的值
-
已知2x-y=1,xy=2,求(xy-5x-2y)-(-3xy-x-4y)=
-
已知x^2+y^2-2x+4y+5=0.求{(x^4-y^4)/[(x+y)(2x-y)]}*{[(2x-y)/(xy-