1-m² 10mn-25n²

(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,

M*N=1/(1/M+1/N)=MN/(M+N)=6得MN=6(M+n)1/M*1/N=1/(M+N)=25得M+N=1/25综上所述MN=6/25

 再答：求采纳！再问：。。。。。

自己的功课自己做

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)=-

再答：原式=2m(n+1)-(n+1)2=2(m-1)(n+1)

你的符号用的不对,m分之一应该是1/m,不是m/1原式=(m^2+n^2)/(m-n)^2-(2/mn)/(1/m-1/n)^2=(m^2+n^2)/(m-n)^2-(2/mn)*[mn/(m-n)]

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

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

已知：a^3-b^3=(a+b)(a^2+ab+b^2)令a=1/5m,b=1/2n则原式==(a+b)(a^2+ab+b^2)=a^3-b^3=(1/5m)^3-(1/2n)^3

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

-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

记得看回答时间复制者死首先求出m与nm²-2m+10=6n-n²m²-2m+10-6n+n²=0(m²-2m+1)+(n²-6n+9)=0（

m^2+n^2=(m+n)^2-2mn=10^2-2x24=100-48=52(m-n)^2=(m+n)^2-4mn=10^2-4x24=100-96=4

原式=3m^2-3mn+6n^2-2mn+2m^2-2n^2+5mn-4n^2+1=5m^2+1再问：（2x-3y+z)-3(x-2(-z+(x-y))-(3y-5z)再答：原式=2x-3y+z-3(

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

就是通分啊n!/(n-m)!=n!(n-m+1)/(n-m+1)!mn!/(n-m+1)!不变相加提出分子公因式n!再问：n!/(n-m)!=n!(n-m+1)/(n-m+1)!不明白怎樣通分麻煩詳解

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

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