若m n=3 求2m² 4mn 2n²-6

由于:mn/(m+n)=2则有：mn=2(m+n)则:原式=(3m+3n-5mn)/(-m-n+3mn)=[3(m+n)-5mn]/[-(m+n)+3mn]=[3(m+n)-10(m

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

3m-5mn+3n/（-m）+3mn-n=【3（m+n）-5mn】/【3mn-（m+n）】=【3-5mn/（m+n）】/【3mn/（m+n）-1】=【3-5x2】/【3x2-1】=-7/5

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

答：mn/(m+n)=2分子分母同除以mn得：1/(1/n+1/m)=21/m+1/n=1/2(3m-5mn+3n)/(-m+3mn-n)分子分母同除以mn得：=(3/n-5+3/m)/(-1/n+3

先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27

由于:mn/(m+n)=2则有：mn=2(m+n)则:原式=(3m+3n-5mn)/(-m-n+3mn)=[3(m+n)-5mn]/[-(m+n)+3mn]=[3(m+n)-10(m+n)]/[-(m

解1/m-1/n=4两边乘以mn得：n-m=4mn∴(2m-3mn-2n)/(m-2mn-n)=[2(m-n)-3mn]/[(m-n)-2mn]=[2×(-4mn)-3mn]/[(-4mn)-2mn]

(-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

（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

-MN(M^2N^5-MN^3-N)=-（-6)^3+(-6)^2-(-6)=258

-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

原式=4m+3mn-4mn+6n=2(2m+3n)-mn=2×6-4=8

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

2（mn-3m)-3（2n-mn）=2mn-6m-6n+3mn=2mn+3mn-6(m+n)=32