已知α²﹢3α+1=O,求α²/α⁴+α²+1的值
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已知α²﹢3α+1=O,求α²/α⁴+α²+1的值
已知α²﹢3α+1=O,求α²/α⁴+α²+1的值
已知α²﹢3α+1=O,求α²/α⁴+α²+1的值
等式两边同除以α
α+3+1/α=0
α+1/α=-3
α²/(α⁴+α²+1)
=1/(α²+1/α²+1)
=1/[(α+1/α)²-1]
=1/[(-3)²-1]
=1/8
a^2+1=-3a
a+1/a=-3
原式=1/(a^2+1/a^2+1)
=1/[(a+1/a)^2-1]
=1/8
α²/α⁴+α²+1 =1/(α²+1+1/α²) 分子分母同除α²
α²﹢3α+1=O可以化为α+1/α =-3 (同时除α)
然后平方得α²1/α² +2 =9
α²1/α²=7
代入上式就是1/8
α²/α⁴+α²+1上下除以α²得到1/(α²+1+1/α²)=1/[(α+1/α)²-1]
由原式除以a可以知道α﹢3+1/α=0,所以α﹢1/α=-3.带入上式为1/8