已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x

已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x
已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x
已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x

已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x已知实数x满足x²-3x+1=0.求（1）x+ 1/x (2)x²+ 1/x² (3)x-1/x
x²-3x+1=0
除以x得 x-3+1/x=0
x+1/x=3
(2)x²+ 1/x²
=x²+2+1/x²-2
=(x+1/x)²-2
=3²-2
=7
(3)(x-1/x)²
=(x+1/x)²-4
=3²-4
=5
∴x-1/x=±√5

x²-3x+1=0
同除以x得
x-3+1/x=0
x+1/x=3
（x+1/x）²=9
x²+1/x²+2=9
x²+1/x²=7
因为x²+1/x²=7
（x²-2+1/x²）+2=7
（x-1/x）²=5
x-1/x=±√5