(m n)²-2(m²-n²) (m-n)²

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,

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

4-m²n²+2mn=5-（m²n²-2mn+1）=5-（mn-1）²=（√5+mn-1）（√5-mn+1）如果本题有什么不明白可以追问,

mn(m-n)-m(n-m)2=m[n(m-n)-(n-m)2]=m(mn-n2-n2+2mn-m2)=m(3mn-2n2-m2)=-m(m2-3mn+2n2)=-m(m-2n)(m-n)另外m的平方

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

(-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

再答：原式=2m(n+1)-(n+1)2=2(m-1)(n+1)

m(m-2n)+n(2mn-m^2)=m(m-2n)+n[m(2n-m)]=m(m-2n)(n+1)

=m^2n^2×(-m^6n^3)=-m^8n^5

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

原式=(m²+2mn+n²-4mn)(m²-2mn+n²+4mn)+2m²n²=(m²-2mn+n²)(m²+

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

2m^2-mn-n^2=(m^2-n^2)+(m^2-mn)=(m+n)(m-n)+m(m-n)=(m-n)(m+n+m)=(m-n)(2m+n)再问：x^2+2x-15谢谢再答：x^2+2x-15=

2[mn+(-3m)]-3(2n-mn)=2mn-6m-6n+3mn=5mn-6m-6n=5mn-6(m+n)m+n=2mn=-3=-15-12=-27

3(m^2n+mn)-4(mn-2m^2n)+mn.=3m^2n+3mn-4mn+8m^2n+mn.=11m^2n.