x^2-4x+1=0 求根号x^2+(1/X^2)-5
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x^2-4x+1=0 求根号x^2+(1/X^2)-5
x^2-4x+1=0 求根号x^2+(1/X^2)-5
x^2-4x+1=0 求根号x^2+(1/X^2)-5
x^2-4x+1=0,两边同除以x得:x+1/x=4;所以:x^2+1/x^2-5=(x+1/x)^2-7=4^2-7=9;所以原式=3.