已知3m-n分之mn
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![已知3m-n分之mn](/uploads/image/f/4205869-61-9.jpg?t=%E5%B7%B2%E7%9F%A53m-n%E5%88%86%E4%B9%8Bmn)
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,
这很简单啊,m=7,n=-4,你算一下就可以了再问:过程了啦...速度再答:∵7+4根号3=m-n根号3所以7=m-(n+4)根号3又因为左边没有无理数所以(n+4)根号3=0所以n=-4所以m=7
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
m/(m-n)-n/(m+n)+2mn/(m²-n²)=[m(m+n)-n(m-n)+2mn]/(m²-n²)=(m²+n²)/(m
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
1/m+2/n=1,1/m=1-2/n=(n-2)/n,m=n/(n-2).n=3时,m=3;n=4时m=2.除此之外,n的其它取值都不能使m为自然数.
-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
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
m分之1-n分之1=3两边乘以mn得n-m=3mn∴m-n=-3mnm-2mn-n分之2m+3mn-2n=[2(m-n)+3mn]/[(m-n)-2mn]=(-6mn+3mn)/(-3mn-2mn)=
1是不是题错了?2由题意有F(0)F(1)0,(-5)(a+2-5)=-5(a-3)0,解得a33分别解方程有m=0或m=3,n=0或n=3.因为MN不相等,所以有(1)M=0,N=3,此时原式=4*
原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8
m分之一减n分之一=5则n-m=5mnm+6mn-n=mn3m+12mn-3n=-3mn答案为-3
m²-2mn-3n²=(m-3n)(m+n)=0m=3n或m=-n所以m/n=3或m/n=-1
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