(x-1/2y-1)（x-1/2y+1）-（x-1/2y-1）的2次
=(x-1)²/[(2y-1)(2y+1)]-(x-1)²/(2y-1)²
=(x-1)²/(2y-1)[1/(2y+1)-1/(2y-1)]
=(x-1)²/(2y-1)(2y-1-2y-1)/[(2y-1)(2y+1)]
=-2(x-1)²/[(2y-1)²(2y+1)]
(x-1/2y-1)（x-1/2y+1）-（x-1/2-1）的2次 简便方法
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