Question

若m-n=4，mn=负1，求代数式（9-2mn+2m+3n）-（3mn+2n-2m）-（m+4n+mn）的值

数学题、要说出过程

Answer 1

解：将mn=-1带入（9-2mn+2m+3n）-（3mn+2n-2m）-（m+4n+mn）所以原式=（9+2+2m+3n）-（-3+2n-2m）-（m+4n-1）去括号=9+2+2m+3n+3-2n+2m-m-4n+1=15+3m-3n

Answer 2

(9-2mn+2m+3n）-（3mn+2n-2m）-（m+4n+mn)把括号展开即可，再合并同类项，所以得结果=-6mn+3m-3n+9=6+4+9=19

Answer 3

(9-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)＝9-2mn+2m+3n-3mn-2n+2m-m-4n-mn=9-6mn+3m-3n
因为m-n=4，mn=-1
9-6mn+3m-3n=9-6mn+3(m-n)=9-6*(-1)+3*4=27