已知x²-3x+1=0,求(1)x²+1/x² (2)x-1/x的绝对值
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已知x²-3x+1=0,求(1)x²+1/x² (2)x-1/x的绝对值
已知x²-3x+1=0,求(1)x²+1/x² (2)x-1/x的绝对值
已知x²-3x+1=0,求(1)x²+1/x² (2)x-1/x的绝对值
很明显x≠0
x²-3x+1=0
两边同除以x得
x-3+1/x=0
x+1/x=3
x^2+1/x^2=(x+1/x)^2-2=7
x-1/x=±√[(x+1/x)^2-4]=±√5