(X) =x2-2x + 2a(a2-2ab-b2)-b(2a2 ab-b2)loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2

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(X) =x2-2x + 2a(a2-2ab-b2)-b(2a2 ab-b2)loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2
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(X) =x2-2x + 2a(a2-2ab-b2)-b(2a2 ab-b2)loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2
(X) =x2-2x + 2a(a2-2ab-b2)-b(2a2 ab-b2)
loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2

(X) =x2-2x + 2a(a2-2ab-b2)-b(2a2 ab-b2)loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2
an=2a(n-1) (n 2)/【n(n 1)】(n≥2,n∈n*)所以a5=8,a7=16,求a1与公比q所以 (X) =x2-2x + 2算式中各项均为向量,下同