原式=2(x⁴+2x²y²+y⁴)÷(x²+y²)
=2(x²+y²)²÷(x²+y²)
=2(x²+y²)
=2x²+2y²
2x^4+4x^2y^2+2y^4除以(x²+y²)
1个回答
相关问题
-
(x^2 +4x+4-y^2)除以(x+y+2)
-
x^2+4y^2-6x+4y+10=0,求x^4-y^4/(x+2y)(x-2y)*x+2y/xy^2+y^3除以（x^
-
8x^2y^4除以(-3x/4y^2)除以(-xy^2/2)
-
y=(2x^2-3x+4)/√ x 求y' y=(2x^2-3x+4)除以根号X 求 y'
-
(4x^2-y^2)^2除以(-2x+y分之4x^2-4xy+y^2)^3 ,计算题
-
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
-
(-3y\x)^3除以(-9x^4y)*(-x^2\y)^4
-
已知x^2+4y^2-2x+4y+5=0,求2x^2+xy-y^2分之x^4-y^4乘以xy-y^2分之2x-y除以(x
-
x-y/x^2+xy除以xy-x^2/x^2y^2-x^4*1/x-y