[x²y²（x－4y）（x＋4y）－x²y²（x-2y）²]／[-（2xy）²]
＝x²y²（x²-16y²-x²＋4xy－4y²）／-4x²y²
＝20y²-4xy／4
＝5y²-xy
[x^2-y^2(x-4y)(x+4y)-x^2-y^2(x-2y)^2]/[-(2xy)^2]
3个回答
相关问题
-
(x^2+4y^2-2xy)(x^2+4y^2+2xy)(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
-
(x^3+y^3)^2-4xy[x^4+x^2y^2+y^4-2xy(x^2-xy+y^2)]因式分解
-
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x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以（x2+y2/x
-
x^2-2xy+y^2=14x^2-4xy+y^2=46x^2-5xy+y^2=0x^2-4xy+3y^2+4x-8y+
-
(x^2+2xy+y^2)-4(x^2-y^2)+4(x-y)^2
-
若x^4+y^4=15,xy^2-x^2y=-5,求x^4-y^4+3xy^2-x^2y-2xy^2+2y^4
-
因式分解：x(y-3)-(2y-6) (x+y)^3-(x-y)^2(x+y) x(x^2-xy)-(4x^2-4xy)
-
（2x+y-1）²=(2x+y)²-2(2x+y)+1=4x²+4xy+y²-4