已知m n=3,mn=2 3
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m+n=2mn=-42(mn-3m)-3(2n-mn)=2mn-6m-6n+3mn=5mn-6(m+n)=-20-12=-32
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,
2(mn+m)-[-(3n-mn)-m]+mn =2mn+2m+3n-mn+m+mn =2mn+3m+2n =2mn+3(m+n) ∵m+n=3,mn=-2 ∴2(mn+m)-[-(3n-m
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
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
m3n-m2n2+mn3,=mn(m2-mn+n2),=mn[(m2+2mn+n2)-3mn],=mn[(m+n)2-3mn],当m+n=3,mn=23时,原式=23(32−3×23)=423.再问:
m²-mn=15==>3m²-3mn=45(1)mn-n²=-6==>2mn-2n²=-12(2)(1)式+(2)式==>3m²-mn-2n²
64=MQ=1/2MP=1/2*4/3MNSOMN=4*2/1*3/4=6
-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
方法一:根据题意,m2-mn=2,mn-n2=5,故有m2=2+mn,n2=mn-5,∴原式=3(2+mm)+2mn-5(mn-5)=31.故应填31.方法二:根据已知条件m2-mn=2,mn-n2=
=(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
∵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
3(2n-mn)+2(mn+3m)=6n-3mn+2mn+6m=6(m+n)-mn=6*-3-2=-20
2(mn-3m)-3(2n-mn)=2mn-6m-6n+3mn=2mn+3mn-6(m+n)=32