(X^2 + 2)(1/X^2 - 1)^5的展开式常数项是
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原方程等价于：
X^2 - 2*(1/2)*X + (1/2)^2 - (1/2)^2 - 1 = 0
=> (X - 1/2)^2 - 5/4 = 0
=> (X - 1/2)^2 = 5/4
=> X - 1/2 = ±√(5/4)
=> X1 = √(5/4) + 1/2
X2 = -√(5/4) + 1/2
=> X1 = (1+√5)/2
=> X2 = (1-√5)/2
最后应该是：X = (1 ± √5)/2
X^2 - 2*(1/2)*X + (1/2)^2 - (1/2)^2 - 1 = 0
=> (X - 1/2)^2 - 5/4 = 0
=> (X - 1/2)^2 = 5/4
=> X - 1/2 = ±√(5/4)
=> X1 = √(5/4) + 1/2
X2 = -√(5/4) + 1/2
=> X1 = (1+√5)/2
=> X2 = (1-√5)/2
最后应该是：X = (1 ± √5)/2
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