Question

先化简再求值:(1-3/x+2)÷x-1/x2+2x-x/x+1.其中x满足x2-x-1=0

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Answer 1

x²-x-1=0
x²-x=1
x²-x+1/4=1+1/4
(x-1/2)²=5/4
x-1/2=±√5/2
x=1/2±√5/2=(1±√5)/2
x²=(1±2√5+5)/2=(6±2√5)/2=3±√5
x³=(1±3√5+15±5√5)/8=(16±8√5)/8=2±√5
（1-3/x+2）÷x-1/x²+2x-x/x+1
=(3-3/x)÷x-1/x²+2x-1+1
=3/x-3/x²-1/x²+2x
=2x+3/x-4/x²
=(2x³+3x-4)/x²
=[2(2±√5)+3(1±√5)/2-4]/(3±√5)
=[4±2√5+(3±3√5)/2-4]/(3±√5)
=[±2√5+(3±3√5)/2]/(3±√5)
=(3±5√5)/2]/(3±√5)
=(3±5√5)(3±√5)/2(3-5)
=-(9±15√5±3√5+25)/4
=-(34±18√5)/4
=-(17±9√5)/2