m-2n=4 2m-n=2

应该是10吧

m-{n-2m+[3m-(6m+3n)+5n]}=m-n+2m-[3m-(6m+3n)+5n]=3m-n-3m+(6m+3n)-5n=-6n+6m+3n=6m-3n

465.据题可知这些元素构成等差数列,且首项a1=1,公差q=2.又因为m

m2=m+2m3=m(n+2)=mn+2mn2=m+2n3=n(m+2)=mn+2nm2-n2=(n+2)-(m+2)(m+n)(m-n)=-(m-n)m≠n则m-n≠0所以m+n=-1原式=mn+2

465.据题可知这些元素构成等差数列,且首项a1=1,公差q=2.又因为m<60,即数列的最大项为59,从而可求出数列一共有n=30项,所以a30=59.那么它们的和S=[n*(a1+a30)]

m(m+n)(m-n)-m(m+n)^2=m(m+n)[(m-n)-(m+n)]=m(m+n)(-2n)=-2mn(m+n)=-2*(-1/2)*1=1

[(m+n)(m-n)-(m-n)²]÷(2n)=[(m+n-m+n)(m-n)]÷(2n)=2n(m-n)÷(2n)=m-n=-2-(-1/2)=-3/2

将m+n看作x,那么算式变成了x的方-6x+9,那么用平方公式一算,就是（x-3）的方,所以答案不就是（m+n-3）的方嘛!至于-3嘛,-3乘以2不就是-6嘛!希望采纳,谢谢!

(m-2n)/(2m+n)=3m-2n=6m+3n5m+5n=0m=-n[3(m-2n)/(2m+n)-(m-2n)/[2(2m+n)]-{9(m-2n)/[4(2m+n)]}=[3(-n-2n)/(

2m(m-n)-(m+2n)(2m-n)=2m^2-2mn-(2m^2+3mn-2n^2)=2m^2-2mn-2m^2-3mn+2n^2=2m^2-2m^2+2n^2-2mn-3mn=2n^2-5mn

m/(m+n)+n/(m-n)-n^2/(m^2-n^2)=[m(m-n)+n(m+n)-n^2]/(m^2-n^2)=m^2/(m^2-n^2)=1/(1-(n/m)^2)=1/(1-(3/2)^2

已知m=5n,则原式=(5n/(5n+n))+(5n/(5n-n))-(n^2)/(((5n)^3)-n^2)=(5/6)+(5/4)-[1/(125n-1)]=(25/12)-[1/(125n-1)

4.5*5=4.5+(4.5+1)+(4.5+2)+(4.5+3)+(4.5+4)+4.5+5)=4.5x6+1+2+3+4+5=27+15=42m*8=37.8m*n=m+(m+1)+(m+2)+(

2m-n/M+2n的绝对值=32m-n/M+2n=3或-32(2m-n)/m+2n减2m-n/m+2n减3=(2m-n/m+2n)-3=0或-6

设m=ta,n=tb=>tb|(2ta-1)=>t|(2ta-1)=>t|1=>t=1所以m,n互素2m=un+1,2n=vm+1相减得(2+v)*m=(2+u)*n由于m,n互素所以m=2+u,n=

=(m^2-n^2)(-m^2-n^2)-(4m^2-n^2)(4m^2+n^2)=-(m^4-n^4)-(16m^4-n^4)=-（1^4-（-2)^4）-(16*1^4-(-2)^4)=15

(2m-n)/(m+2n)=5∴(m+2n)/(2m-n)=1/5∴3(2m-n）/m+2n-m+2n/2m-n+3=3×(2m-n)/(m+2n)+(m+2n)/(2m-n)+3=15+1/5+3=

错排公式中的n是指的错排数,所以把n-k替换原公式中的n就行了M(n-k)=(n-k-1)[M(n-k-2)+M(n-k-1)]

依题意,M={1,3,5,..,59},共有30个这些元素的和=1+3+..+59=30^2=900

2x+n/m=m/n-(m/n+n/m)x(m/n+n/m+2)x=m/n-n/m﹙m+n﹚²x／﹙mn﹚=﹙m+n)(m-n)／﹙mn﹚∵m+n不等于0∴x=(m-n)／﹙m+n﹚