2道初二分式计算1.a^(n+1)-6a^n+9a^(n-1)2.a^(n+1)-4a^n+4a^(n-1)这两个怎么分
2道初二分式计算1.a^(n+1)-6a^n+9a^(n-1)2.a^(n+1)-4a^n+4a^(n-1)这两个怎么分
分式 计算:a^(2n+1)-6a^(2n)+9a^(2n-1) / a^(n+1)-4a^n+3a^(n-1)
计算:(3a^n+2+6a^n+1-9^n)÷3a^n-1
证明n^n-n(n-a)^(n-1)>=n!a.其中n>=a>0
计算:(6a^n+2-9a^n+1+3a^n-1)÷3a^n-1
(3a^(n+1)+6a^(n+2)-9a^n)/3a^(n-1)是计算
计算a^2n+1-6a^2n+9a^2n-1/a^2-9除以a^n+1+4a^n+4a^n-1/a^2-4
a^(n+1)b^n-4a^(n+2)+3ab^n-12a^2
(3a^n-2)-6a^n+14a^n-1(因式分解) (m-n)^3+4(n-m)
分式计算:(b^3n-1 )*c/(a^2n+1)除以 (b^3n-2)/(a^2n)
计算:(3a^(n+1)-6a^n+9a^(n-1)/(1/3a^(n-1)
A(n,n)+A(n-1,n-1)=XA(n+1,n+1)