若2m n=25,m-2
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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
原式=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=3(m-n)-6mn=3×2-6×1=0.
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
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
=[m(m+n)/(m^2-n^2)-n(m-n)/(m^2-n^2)+2mn/(m^2-n^2)]÷(m^2+mn)/m=(m^2+mn-mn+n^2+2mn)/(m+n)(m-n)÷(m+n)=(
先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27
解题思路:将原代数式经过去括号后合并同类项,化简后代入数值计算即可。解题过程:
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)=-
(9-2mn+2m+3n)-(m+4n+mn)=9-2mn+2m+3m-m-4n-mn=9-mn+(m-n)=9-(-1)+4=14
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
原式=(m^2+n^2-2mn)/2=(m-n)^2/2=(-2)^2/2=4/2=2如果你对我的回答满意,请【采纳为满意答案】,可继续询问,直至弄懂
(0,根号n-m,0)(0,-根号n-m)
-MN(M^2N^5-MN^3-N)=-(-6)^3+(-6)^2-(-6)=258
∵m+n=5,mn=-6,∴(m+n)2 −4mn=52-4×(-6)=25+24=49,故答案为:49.
-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
2[mn+(-3m)]-3(2n-mn)=2mn-6m-6n+3mn=5mn-6m-6n=5mn-6(m+n)m+n=2mn=-3=-15-12=-27
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
(N-M)²-4MN=5²-4*(-2)=25+8=33
若m^2+n^2=25,mn=12因为(m+n)^2=m^2+n^2+2mn=25+24=49所以m+n=±7