已知a^2+4a+1=0,求(1)a^2+1/a^2(2)(a-2)(1/a+2)(3)a^4+1/a^4

题目

已知a^2+4a+1=0,求(1)a^2+1/a^2(2)(a-2)(1/a+2)(3)a^4+1/a^4
求解!

答案

已知a^2+4a+1=0,
a+1/a=-4;
求(1)a^2+1/a^2
=(a+1/a)²-2
=16-2
=14;
(2)(a-2)(1/a+2)
=1+2a-2/a-4
=-3+2(a-1/a)
=-3+2×±√[(a+1/a)²-4]
=-3±2×2√3
=-3+4√3或-3-4√3；
(3)a^4+1/a^4
=(a²+1/a²)²-2
=14²-2
=196-2
=194;
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