若m-n=4,mn=-1,求代数式(9-2mn 2m 3n)

(2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)先去括号=2mn+2m+3n-3mn-2n+2m-m-4n-mn合并同类项=-2mn+3m-3n=-2mn+3(m-n)把m-n=2,

（9-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=9-2mn+2m+3n-3mn-2n+2m-m-4n-mn=9-6mn+3m-3n=9-6mn+3(m-n)=9-6×（-1）+

（-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），又因为m-n=4，mn=-

(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=(-2mn-3mn-mn)+(2m+2m-m)+(3n-2n-4n)=-

(1)已知m-n=2,mn=1所以（-2mn+2m+3n）-（3mn+2n-2m）-（m+4n+mn）=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=-6mn+3m-3n=-6mn+3(m

解（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn）=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=-2mn-3mn-mn+2m+2m-m+3n-2n-4n=-6mn+3

先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27

(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=(-2mn-3mn-mn)+(2m+2m-m)+(3n-2n-4n)=-

（9-2mn+2m+3n)-(m+4n+mn）=9-2mn+2m+3m-m-4n-mn=9-mn+(m-n)=9-(-1)+4=14

∵丨m-n-4丨≥0,（mn+1）²≥0又丨m-n-4丨+（mn+1）²=0∴丨m-n-4丨=0,（mn+1）²=0∴m-n-4=0,mn+1=0∴m-n=4①,mn=-

=-2MN+2M+3N-3MN-2N+2M-M-4N-MN=-2MN-3MN-MN+3M-3N=-6MN+3(M-N)已知M-N=3,MN=1所以是-6*3+3*1=-15

-5m²n+（2mn²+5m²n）-2（3m²-mn²)=-5m²n+2mn²+5m²n-6m²+2mn

（4m-5n-mn)-(2m-3n+5mn)=4m-5n-mn-2m+3n-5mn=2m-2n-6mn=2(m-n)-6mn当m-n=2,mn=-1时,原式=2×2-6×（-1）=10已知A=2x^2

n/m+4m/n-1=(n^2+4m^2-mn)/mn=((2m-n)^2+3mn)/mn=(3mn+3mn)/mn=6

-2mn+2m+3n-3mn-2n+2m-4n-m-mn=-6mn+3m-3n=-6mn+3(m-n)=6+9=15

原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8

4m²-4mn+n²/2m+n除以（4m²-n²）约分等于（2m-n）/【(2m+n)^2】因为4m²+n²=5,mn=-1,所以4m

（-2mn+2m+3n）－（3mn+2n-2m）－（m+4n+mn）=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=(-2mn-3mn-mn)+(2m+2m-m)+(3n-2n-4n)=-

已知mn=-1,m-n=4则(-2mn+m+n)-(3mn+5n-5m)-(m+4n-3mn)=-2mn+m+n-3mn-5n+5m-m-4n+3mn=-2mn+5m-8n=2+20-3n=22-3n

(-m-4n-mn)-(2mn-2m-3n)-(3mn+2n-2m)=-m-4n-mn-2mn+2m+3n-3mn-2n+2m=3m-3n-6mn=3(m-n)-6mn=3×3-6×(-3)=9+18