Question

已知m-n=2,mn=1,求多项式(-2mn+2mn+3n)-(3mn+2n-2m)-(m+4n+mn)的值

已知m-n=2,mn=1,求多项式(-2mn+2mn+3n)-(3mn+2n-2m)-(m+4n+mn)的值

Answer 1

错了，改一下
(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=-6mn+3m-3n
=-6mn+3(m-n)
=-6+6
=0

Answer 2

解:化简(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2mn+3n-3mn-2n+2m-m-4n-mn
=-6mn+3m-3n
=-6×1+3(m-n)
=-6+3×2
=0