关于x y的多项式6mx²+4nxy+2x+2xy-x²+y+4不含二次项 求多

问题补充：

关于x,y的多项式6mx²+4nxy+2x+2xy-x²+y+4不含二次项,求多项式2m²n+10m-4n2+2m²n+4m+2n的值

答案：

6mx²+4nxy+2x+2xy-x²+y+4=(6m-1)x²+(4n+2)xy+2x+y+46m-1=0m=1/64n+2=0n=-1/22m²n+10m-4n2+2m²n+4m+2n=4m²+14m-4n²+2n=4x(1/6)²+14x1/6-4x(-1/2)²-2x1/2=1/9+7/3-1-1=1...

======以下答案可供参考======

供参考答案1:

6mx²+4nxy+2x+2xy-x²+y+4

=(6m-1)x²+(4n+2)xy+2x+y+4 不含二次项，则 6m-1=0，4n+2=0

m=1/6，n= -1/2

2m²n+10m-4n2+2m²n+4m+2n

=4m²n+14m-4n^2+2n

=2m(2mn+7)-2n(2n-1)

=5/18供参考答案2:

由题可知6mx²+4nxy+2x+2xy-x²+y+4

=(6m-1)x²+(4n+2)xy+2x+y+4即6m-1=0,4n+2=0 可得 m=1/6,n=-1/2

所以代入2m²n+10m-4n2+2m²n+4m+2n可得其值