计算:(3a^(n+1)-6a^n+9a^(n-1

计算:(3a^(n+1)-6a^n+9a^(n-1)/(1/3a^(n-1)

2个回答

(3a^(n+1)-6a^n+9a^(n-1)/(1/3a^(n-1)
=3[3a^(n+1)-6a^n+9a^(n-1)]/a^(n-1)
=3[3an^(n-1)(a^2-2a+3)]/a^(n-1)
=3*3*(a^2-2a+3)
=9a²-18a+27

计算(3a^n+1-6a^n+9a^n-1)/(1/3a^n-2)
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（6a^n+2+3a^n+1-9a^n)÷1/2a^n-1
-3a^n(a^n-1+2a^n-2+3a^n-3)+a^n-2(a^n-1-a^n+4a^n+1)
{[a^(2n+1)-6a^(2n)+9a^(2n-1)]/(a^2-9)}/[a^(n+1)+4a^(n)+4a^(n
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计算(3a的n+1次方-6a的n次方+9a的n-1次方)除以(3分之1a的n-2次方)