[m-n][m n] [m n]的平方-2m的平方.其中mn满足方程组-搜答案-灵鹊做题网

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,

原式=（2m-m）/m=m/m=1~

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

(-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)=-

因为m-mn=21,mn-n=15,所以：m-n=(m-mn)+(mn-n)=21+15=36m-2mn+n=(m-mn)-(mn-n)=21-15=6希望能都帮到你,追问：对不起啊.我把题发错了,m

m(m-n)(n+1)

-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

mn+m^2n^2+m^2n+mn^2=m^2n(n+1)+mn(n+1)=mn(n+1)(m+1)或：mn+m^2n^2+m^2n+mn^2=mn(mn+m+n+1)=mn(n+1)(m+1)

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

mn(m-n)-(n-m)=mn(m-n)+(m-n)=(mn+1)(m-n)

m+n=-2mn=-4则2(mn-3m)-3(2n-mn)的值=2mn-6m-6n+3mn=5mn-6(m+n)=5*(-4)-6(-2)=-20+12=-8

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

m的平方n-mn的平方

m²n-mn²=mn(m-n)

=(m^2-mn)+2(m^2-n^2)＝(m^2-mn)+2(m^2-mn)+2(mn-n^2)题目条件打错了,自己代入一下

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

∵原式=-3(2n-mn)+2(mn-3m)=-6(m+n)+5mn∵m+n=-3,mn=2∴原式=-6·-3+5·2=28

(-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