已知 m−n=−2 ,则−m n= ,4−2m 2n=
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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
m³n+mn³=mn(m²+n²)=-2x1=-2请好评~在右上角点击【评价】,然后就可以选择【满意,问题已经完美解决】了.如果你认可我的回答,敬请及时采纳,~
(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^2-2mn+n^2+mn(前面多减一个配成完全平方,后面再加一个)=(m-n)^2+mn=(m+n)^2+5mn=81+70=151
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
2m²+4mn+2n²-6=2(m+n)²-6=12
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
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
-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
m²+(n-mn)²+2mn-2m²n=m²+n²-2mn²+m²n²+2mn-2m²n=m²+2m
∵m+n=5,mn=-6,∴(m+n)2 −4mn=52-4×(-6)=25+24=49,故答案为:49.
原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8
(m-n)²-(m+n)²=[(m-n)+(m+n)][(m-n)-(m+n)]=(2m)*(-2n)=-4mn=-2
知m-n=7,mn=2求5n-3mn+5m+2(m+n)²=(m-n)²+4mn=49+8=57m+n=正负根号575n-3mn+5m+2=5n+5m-3mn+2=5(m+n)-3
(-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
m²+2m+n²-6n+10=(m+1)²+(n-3)²=0等式成立,只有m+1=0,n-3=0解得m=-1n=3所以mn=-3