2m-n=2n 3m-2n=5
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1.m=2^n+5n-2在n∈N*上单调递增,所以当n=1时,m=5
既然是相反数,就是m和n两者加起来为0的,例如-1和1,-2和2就互为相反数,所以这里答案是0咯
3m+2n=7(1)3m-n=5(2)(1)-(2)得:2n+n=7-5=2;n=2/3;带入(1)得:3m+4/3=7;3m=17/3;m=17/9;如果本题有什么不明白可以追问,
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
解题思路:给出的算式中,除式和被除式相同,那么商为1解题过程:解:
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
m+n=5m-n=6则mn^2-m^2n=mn(n-m)(m+n)^2=m^2+n^2+2mn=25(m-n)^2=m^2+n^2-2mn=364mn=25-36=-11mn=-11/4mn^2-m^
3m=5n+13m-6n=-n+1n=3(2n-m)+1=3(2n-m+1)-2所以可令n=3k+1或者3k-2.
答:(5m+3n)^2-(m-3n)(25m-3n),其中m=1,n=2=25m²+30mn+9n²-(25m²-3mn-75mn+9n²)=30mn+78mn
(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)/(
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)
2m=5nn=2m/5则把n带入所求的式子得答案7/5~~~~很简单的,我没有仔细算,过程对的,答案你再算一遍楼下的不对~~很明显是n/m-m/(m-n)你认为如果是m/m的话有必要写上去么~
=(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=
(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=
2m+n分之m-2n=3所以原式=(3+1/3-5/4)[2m+n分之m-2n=3]=25/12×3=25/4