
3个回答
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3 = x + 1/x. x不为0。
若x<0,则x + 1/x<0与3 = x+1/x矛盾。
因此,x>0.
0<x<1时,0<x<1<1/x. x - 1/x < 0.
9 = (x+1/x)^2 = x^2 + 1/x^2 + 2,
5 = x^2 + 1/x^2 - 2 = (x- 1/x)^2,
-5^(1/2) = x - 1/x.
x^2 - 1/x^2 = (x+1/x)(x-1/x) = 3[-5^(1/2)] = -3*5^(1/2)
x>1时,x>1>1/x>0, x - 1/x >0.
9 = (x+1/x)^2 = x^2 + 1/x^2 + 2,
5 = x^2 + 1/x^2 - 2 = (x-1/x)^2,
5^(1/2) = x - 1/x,
x^2 - 1/x^2 = (x+1/x)(x-1/x) = 3*5^(1/2).
综合,有,
0<x<1时,x^2 - 1/x^2 = -3*5^(1/2).
x>1时,x^2 - 1/x^2 = 3*5^(1/2)
若x<0,则x + 1/x<0与3 = x+1/x矛盾。
因此,x>0.
0<x<1时,0<x<1<1/x. x - 1/x < 0.
9 = (x+1/x)^2 = x^2 + 1/x^2 + 2,
5 = x^2 + 1/x^2 - 2 = (x- 1/x)^2,
-5^(1/2) = x - 1/x.
x^2 - 1/x^2 = (x+1/x)(x-1/x) = 3[-5^(1/2)] = -3*5^(1/2)
x>1时,x>1>1/x>0, x - 1/x >0.
9 = (x+1/x)^2 = x^2 + 1/x^2 + 2,
5 = x^2 + 1/x^2 - 2 = (x-1/x)^2,
5^(1/2) = x - 1/x,
x^2 - 1/x^2 = (x+1/x)(x-1/x) = 3*5^(1/2).
综合,有,
0<x<1时,x^2 - 1/x^2 = -3*5^(1/2).
x>1时,x^2 - 1/x^2 = 3*5^(1/2)
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解:x+1/x=3两边平方得x^2+1/x^2+2=9,∴x^2+1/x^2=7
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