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因式分解;a^4x^n+2-4x^n (n为正整数)=多少因式分解;25a^n+2-10a^n+1+a^n=多少?
人气:117 ℃ 时间:2020-06-14 21:09:57
解答
a^4x^n+2-4x^2=(a^4*x^2-4)*x^n=(a^2*x+2)(a^2*x-2)*x^n
25a^n+2-10a^n+1+a^n=(25a^2-10a+1)*a^n=(5a-1)^2*a^n
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