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
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![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](/uploads/image/z/2555736-24-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%E6%BB%A1%E8%B6%B3x%26%23178%3B-3x%2B1%3D0.%E6%B1%82%EF%BC%881%EF%BC%89x%2B+1%2Fx+%282%29x%26%23178%3B%2B+1%2Fx%26%23178%3B+%283%29x-1%2Fx%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%E6%BB%A1%E8%B6%B3x%26%23178%3B-3x%2B1%3D0.%E6%B1%82%EF%BC%881%EF%BC%89x%2B+1%2Fx+%282%29x%26%23178%3B%2B+1%2Fx%26%23178%3B++%283%29x-1%2Fx)
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