Question

因式分解以下两题，大神求教了！

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Answer 1

原式
=（a²+9a²-1+6ab）（a²+9a²-1-6ab）
={a+3b）²-1}{（a-3b）²-1}
=(a+3b+1)(a+3b-1)(a-3b+1)(a-3b-1)
原式
=（x^n+1/2）²-1/9y^4m
=(x^n+1/2+1/3y^2m)(x^n+1/2-1/3y^2m)

Answer 2

1、原式=(a^2+9b^2-1+6ab)(a^2+9b^2-1-6ab)
=[(a+3b)^2-1][(a-3b)^2-1]
=(a+3b+1)(a+3b-1)(a-3b+1)(a-3b-1)
2、原式=(x^n+1/2)^2-(1/9)*y^(4m)
=(x^n+1/2+(1/3)*y^2m)(x^n+1/2-(1/3)*y^2m)

Answer 3

(1)
=(a^2+9b^2-1+6ab)(a^2+9b^2-1-6ab)
=[(a+3b)^2-1][(a-3b)^2-1]
=(a+3b+1)(a+3b-1)(a-3b+1)(a-3b-1)
(2)
=(x^n+1/2)^2-(1/9)y^4m
=(x^n+1/2+y^2m/3)(x^n+1/2-y^2m/3)

Answer 4

见图