Question

我有一道初一数学题，请大家帮帮我，急

已知m-n=4,mn=-1,求（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值

Answer 1

解：原式=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=3m-3n-6mn
=3(m-n)-6mn
∵m-n=4 mn=-1
∴3(m-n)-6mn=3*4-6*(-1)=18

Answer 2

去括号得-2mn+2m+3n-3mn-2n+2m-m-4n-mn,合并同类项得-6mn-3n+3m=18

Answer 3

（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=-6mn+3m-3n
=-6*(-1)+3*4
=18

Answer 4

去括号，化简得
原式=-6mn+3m-3n
=-6mn+3（m-n)
=6+12
=18

Answer 5

（-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*(-1)+3*4
=18