3.m n=19 (1)m.n均为质数 (2)5m 7n=129
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(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,
mn/(3m-n)=3,则:(3m-n)/(mn)=1/3,则:2mm/(3m-n)+(6m-2n)/(mn)=2[(mn)/(3m-n)]+2×[(3m-n)/(mn)]=2×3+2×(1/3)=2
先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27
不妨设m≤n,由mn|m²+n²得m|m²+n²,故m|n²,m|n,设n=km,有mn=km²,m²+n²=k
(-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)=-
(1-M)(1-N)=1-M-N-MN=1-(M+N)-MN=1-2-(-2)=1
m-mn+n=﹙m-2mn+n﹚=﹙m-n﹚=×2=×4=2
(m+n)²-4mn=m^2+2mn+n^2-4mn=m^2-2mn+n^2=(m-n)^2=(1/2)^2=1/4
m²n²+m²+n²+1-4mn=0(m²n²-2mn+1)+(m²-2mn+n²)=0(mn-1)²+(m-
m(m-n)-n(n-m)=m(m-n)+n(m-n)=(m+n)(m-n)=12=2X2X3=2X6=(4-2)X(4+2)所以m=4,n=2再问:2*2*3为何等于2*6,不等于4*3,可以说清楚
-2mn+2m+3n-3mn-2n+2m-4n-m-mn=-6mn+3m-3n=-6mn+3(m-n)=6+9=15
(m-1)(n-1)=mn-m-n+1=mn-(m+n)+1=1-2+1=0再问:再问一道题:已知对整式p=(x-2)(x-3),Q=(x+1)(x-6),如果其中x取值相同时,整式p与Q的关系为直接
由式1可以得出(mn-1)^2+(m-n)^2=0,这样可以得出m=n=1.其实这种题可以直接假设m=1,看看n等于多少,因为求答案,没错的
m+n=-2mn=-4则2(mn-3m)-3(2n-mn)的值=2mn-6m-6n+3mn=5mn-6(m+n)=5*(-4)-6(-2)=-20+12=-8
原式=[1/n+1-1/m]/[3/n-3-3/m]=[-(1/m-1/n)+1]/[-3(1/m-1/n)-3]=(-3+1)/(-9-3)=(-2)/(-12)=1/6
-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
m>n>0,m^2+n^2=3mn,——》(m+n)^2=m^2+n^2+2mn=5mn,——》m+n=√(5mn),(m-n)^2=m^2+n^2-2mn=mn,——》m-n=√(mn),——》(m
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
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