Question

已知x²-3x+1=0时,求根号下x²+1/x²-2

求根号下x²+(1/x²)-2

Answer 1

已知x²-3x+1=0
求√[(x²+1)/(x²-2)]的值
解：由x²-3x+1=0，得x²+1=3x；x²-2=3x-3；故√[(x²+1)/(x²-2)]=√[3x/(3x-3)]=√[x/(x-1)]........(1)
再有x²-3x+1=0，得x=(3±√5)/2；
当x=(3+√5)/2时，代入(1)式得√[(x²+1)/(x²-2)]=√[(3+√5)/(1+√5)]
=(1/2)√[(3+√5)(√5-1)]=(1/2)√(2+2√5)
当x=(3-√5)/2时，代入(1)式得√[(x²+1)/(x²-2)]=√[((3-√5)/(1-√5)](虚数，舍去)
故√[(x²+1)/(x²-2)]=(1/2)√(2+2√5).