(1-3/x+2)÷x-1/x²+2x-x/x+1,其中x满足x²-x-1=0
1个回答

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