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若a是方程x²+3x-1=0的根,求a²+1/a²
人气:141 ℃ 时间:2019-10-27 08:31:21
解答
a²+1/a²=(a^4+1)/a²
=(a²(a²+3a-1)-3a^3+a²+1)/a²
=(a²(a²+3a-1)-3a(a²+3a-1)+9a²-3a+a²+1)/a²
=(a²(a²+3a-1)-3a(a²+3a-1)+11a²-(a²+3a-1))/a²
=(0-0+11a²-0)/a²
=11
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