Question

已知x²-3x-1=0,求①x²+1/x²;②x²-1/x²

快……急需……

Answer 1

x²-3x-1=0
乘以x得 x-3-1/x=0
∴x-1/x=3
x²+1/x²
=(x-1/x)²+2
=3²+2
=11
(x+1/x)²
=(x-1/x)²+4
=3²+4
=13
x+1/x=±√13
x²-1/x²
=(x+1/x)(x-1/x)
=3×﹙±√13﹚
=±3√13

Answer 2

x² - 3x - 1 = 0
x - 1/x = 3

(x - 1/x)² = 9
x² + 1/x² = 11

(x + 1/x)² = x² + 1/x² + 2 = 13

①
x² + 1/x² = 11

②
(x + 1/x) = ±√13
x - 1/x = 3

x² - 1/x²
= (x + 1/x)(x - 1/x)
= ±√13

Answer 3

已知x²-3x-1=0,
x-3-1/x=0
x-1/x=3
x^2-2+1/x^2=9
x^2+1/x^2=11
x^2+2+1/x^2=11+2
(x+1/x)^2=13
x+1/x=±√13
x²-1/x²
=(x-1/x)(x+1/x)
=±3√13

Answer 4

x²-3x-1=0
乘以x得 x-3-1/x=0
∴x-1/x=3
x²+1/x²
=(x-1/x)²+2
=3²+2
=11
(x+1/x)²
=(x-1/x)²+4
=3²+4
=13
x+1/x=±√13
x²-1/x²
=(x+1/x)(x-1/x)
=3×﹙±√13﹚
=±3√13
你是咋学的呀，明显的完全平方公式的应用吗