若m n=2分之1,求(1)3m 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=(-2mn-3mn-mn)+(2m+2m-m)+(3n-2n-4n)=-

3x^2-5x/2+11=x-m/2-3x+n-m/2-3x=-5x/2m=-xn=3x^2+113x^2-(5x+11)/(5x-2)=m/(x-2)-n/(3x+1)(15x^3-6x^2-5x-

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

原式=(3m-n)²分之m(3m-n)=(3m-n)分之m=(3*1-2)分之1=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)=-

由1/m-1/n=2得,n-m=2mn所以(3m+mn-3n)/(m-mn-n)=（mn-3*2mn)/(-mn-mn)=(-5mn)/(-2mn)=5/2

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

2m²-3m+1=0m=[3±√(3²-4*2)]/4=[3±1]/4m=1/2m=1n²+n-2=0(n+2)(n-1)=0n=-2n=1∵,mn不等于1∴m=1/2n

解2(-mn+3n+4m)-1/2(14n+18m-mn)=(-2mn+1/2mn)+(6n-7n)+(8m-9m)=-3/2mn-n-m=-3/2mn-(n+m)=-3/2×(-2)-3=3-3=0

m-mn+n=﹙m-2mn+n﹚=﹙m-n﹚=×2=×4=2

（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

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

根据题意绝对值和完全平方非负所以mn-1=0m-n-2=0mn=1m-n=2（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-m

m分之1-n分之1=3两边乘以mn得n-m=3mn∴m-n=-3mnm-2mn-n分之2m+3mn-2n=[2(m-n)+3mn]/[(m-n)-2mn]=(-6mn+3mn)/(-3mn-2mn)=

-2(mn-3m平方)-{m平方-5(mn-m平方)+2mn},其中m=1,n=-2=-2mn+6m^2-m^2+5(mn-m^2)-2mn=-2mn+6m^2-m^2+5mn-5m^2-2mn=mn

1是不是题错了?2由题意有F(0)F(1)0,(-5)(a+2-5)=-5（a-3）0,解得a33分别解方程有m=0或m=3,n=0或n=3.因为MN不相等,所以有（1）M=0,N=3,此时原式=4*

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