A(1,m),B(n, 3),AOB=45

原式=(a-b)[(m+n)-3(m-n)]=(a-b)(-2m+4n)=2(a-b)(2n-m)

原式=a^mb^2n[a^(-1)b-ab^(-1)]=a^(m-1)b^(2n-1)[b²-a²]=a^(m-1)b^(2n-1)(a+b)(b-a)找好公因式a^(m-1)b^

1、=〔m(m-n)+m(m+n)-n^2〕/(m^2-n^2）=(m^2-mn+m^2+mn-n^2)/(m^2-n^2）=〔m^2/(m^2-n^2)〕+1=〔1/(m^2-n^2)/m^2〕+1

解题思路：考察长方形的性质：长方形的对边相等，长方形的周长等于四边之和解题过程：解：因为长方形的对边相等所以另一边的长=[4M-2(M-N)]/2=M+N选C最终答案：C

7(a-1)+x(a-1)=(7+x)(a-1)3(a-b)²+6(a-b)=(a-b)(3(a-b)+6)=(a-b)(3a-3b+6)2(m-n)²-m(m-n)=(m-n)(

a^n+b^n-a^mb^(n-m)-a^(n-m)b^m=a^m(a^(n-m)-b^(n-m))-(a^(n-m)-b^(n-m))b^m=(a^m-b^m)(a^(n-m)-b^(n-m))1)

因为A和B关于Y轴对称,所以纵坐标相同,即2m+n+5=-m+2n-3,即3m-n+8=0,横坐标互为相反数,即3m-2=-（m-n-1）,即2m-n-3=0,将求出的两个等式联力,即将3m-n+8=

a-ab/a+b（a/a-b+b/b-a）=a-b+b(1-b+1-a)=a-b+2b-b^2-ab=a+b-ab+b^21-m-n/m+2n÷m²-n²/m²+4mn+

A>B因为：A=（a+1）/（a^2+1）,B=（a^2+1）/（a^3+1）所以：A-B=（a+1）/（a^2+1）-（a^2+1）/（a^3+1）=【（a+1）（a^3+1）-（a^2+1）^2】

a^m+1b^n+a^mb^n+1=a^m(1+b^n)+(b^n+1)=(a^m+1)(b^n+1)

=(a^m+1)^3(b^n)^3(c^2)^3=a^(3m+3(b^3n)(c^6)

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

[6a^(m+2)b^(n+2)-4a^(m+1)b^(+1)+2a^(m)b^m]÷(-2a^(m)b(n)=2a^m[3a^2b^(n+2)-2a^b^(+1)+b^m]÷(-2a^(m)b(n)

(a^m+1b^n+2)(a^2n-1b^2m)=a^5b^3,求m-3n的值(a^m+1b^n+2)(a^2n-1b^2m)=a^5b^3a^(m+1+2n-1)b^(n+2+2m)=a^5b^3∴

解析：因为a^(m+1)b^(n+2)·a^（2n-1）·b=a^(m+1+2n-1)·b^(n+2+1)=a^(m+2n)·b^(n+3)=a^5b^3所以可得：n+3=3,m+2n=5解得n=0,

应该是这样的吧:1/27a^(3m-n)-1/3a^(m-n)*b^2n=1/27a^(m-n)[a^(2m)-9b^(2n)]=1/27a^(m-n)*(a^m+3b^n)(a^m-3b^n)

a^(m+1)b^n-a^(m-1)b^(n+2)=a^(m-1)·b^n·(a²-b²)=a^(m-1)·b^n·(a+b)(a-b)

m-1=n-2mm+n=3m+n-4解出m,n,带到方程组里就好了大概是y=20,x=90吧

(1/2a^m-b^n)(b^n+1/2a^m)=(1/2a^m)²-(b^n)²=a^2m/4-b^2n

m+1+2n-1=5;n+2+1=3;得：n=0;m=5