4^m=a,8^n=b,求2^2m 3n
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1.a^3m+b^2n=(a^m)^3+(b^n)^2=3^3+4^2=432.(3x^3n)^2-4(x^2)^2n=9x^6n-4x^4n=3087-196=2891
解根据题意得2m=3m-1n+6=2m+1∴m=1n=-3再问:拜托是8*a^3m-1*b^2m+1再答:��ѽ��Ҫͬ���˵��a��ָ����ͬ��b��ָ��Ҳ��ͬ再问:可是有乘8呀再答:
解(a^m+1b^n+2)·(a^2n-1b^2m)=a^(m+2n)b^(n+2+2m)所以m+2n=5n+2m+2=9解得m=3,n=1所以m^(2n)=3^2=9
由A=-N+M次根号M+N+10是M+N+10的算术平方根得:M-N=2由B=-2N+M+3次根号4M+6N-1的立方根求B-A的立方根得:M-2N+3=3解得:M=4,N=2则A=M-N次根号M+N
(1)8^m+n=2^3^(m+n)=(2^m*2^n)^3=a^3b^32^m+n十2^3m+2n=(2^m*2^n)+(2^m^3)*(2^n^2)=ab+a^3b^2
这种称为讨论题:1)向量a=b=0则n,m∈R2)a=0b不等于0时-2mb+nb=5b-2m+n=5,m=∈R,n=(5+2m)3)a不等于0b=0时3ma+4na=2a;3m+4n=2;m∈R,n
(a^2m)^3+(b^n)^3-a^2m×b^n×a^4m×b^2n=a^6m+b^3n-a^6m×b^3n=(a^3m)^2+b^3n-(a^3m)^2×b^3n=3^2+2-3^2×2=-7
设m+1=an-1=b(m+1)*(n-1)=a*b=(a+1)(b+4)-(a+2)(b+3)=(m+1+1)(n-1+4)-(m+1+2)(n-1+3)=(m+2)(n+3)-(m+3)(n+2)
a^3m=3则a^2m=9的立方根b^3n=2b^n=2的立方根(a^2m)^3+(b^n)^3-a^2mb^n=a^6m+b^3n-a^2mb^n=(a^3m)^2+b^3n-a^2mb^n=9+2
两个等式左右分别相加2a+2b+2(m+n)=35+48化简2*(a+b+m+n)=83所以结果为83/2即41.5
方程整理得,a^m+2nb^2m+n+2=a^5b^3,则有,m+2n=5;2m+n+2=3;解得m=-1,n=3,故,m-3n=-1-9=-10
a^2+9b^2+6a-6b+10=0a^2+6a+9+9b^2-6b+1=0(a+3)^2+(3b-1)^2=0(a+3)^2=0,(3b-1)^2=0a=-3b=1/3[6a^(m+2)b^(n+
(a^2m)^3+(b^n)^3-a^2m×b^n×a^4m×b^2n=a^6m+b^3n-a^6m*b^3n=(a^3m)^2+b^3n-(a^3m)^2*b^3n=9+2-9*2-7
a^(m+1)b^(n+2)*a^(2n-1)*b=a^5b^4a^(m+1+2n-1)b^(n+2+1)=a^5b^4a^(m+2n)b^(n+3)=a^5b^4n+3=4n=1m+2n=5m+2=
2^(3m+10n)=2^3m×2^10n=(2^m)³×[(2^5)^n]²=(2^m)³×(32^n)²=a³×b²=a³b&
由题可知,原式=(m+2)(n+3)-(m+3)(n+2)=m-n,所以当m=2012,n=2011时,原式=1.
解a^m=4a^n=16——这个错了吧a^(2m+2n)=a^2m×a^2n=(a^m)²×(a^n)²=16+256=272
(a^2m)^3+(b^n)^3-a^2m×b^n×a^4m×b^2n=a^6m+b^3n-a^6m×b^3n=(a^3m)^2+b^3n-(a^3m^2×b^3n)=3^2+2-(3^2×2)=9+
m,n共线,m=tna=m+2n=(t+2)n,a,n共线b=3m-4n=(3t-4)n,b,n共线a与b共线