化简：(x^2+1/x^2-3x - x^2-1/x^2-2x-3)除x+1/x
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解：原式=[(x²+1)/(x²-3x)-(x²-1)/(x²-2x-3)]÷[(x+1)/x]
={(x²+1)/[x(x-3)]-(x+1)(x-1)/[(x+1)(x-3)]}×[x/(x+1)]
={(x²+1)/[x(x-3)]-(x-1)/(x-3)}×[x/(x+1)]
={(x²+1)/[x(x-3)]-x(x-1)/[x(x-3)]}×[x/(x+1)]
={[(x²+1)-x(x-1)]/[x(x-3)]}×[x/(x+1)]
={(x²+1-x²+x)/[x(x-3)]}×[x/(x+1)]
={(x+1)/[x(x-3)]}×[x/(x+1)]
=1/(x-3)
={(x²+1)/[x(x-3)]-(x+1)(x-1)/[(x+1)(x-3)]}×[x/(x+1)]
={(x²+1)/[x(x-3)]-(x-1)/(x-3)}×[x/(x+1)]
={(x²+1)/[x(x-3)]-x(x-1)/[x(x-3)]}×[x/(x+1)]
={[(x²+1)-x(x-1)]/[x(x-3)]}×[x/(x+1)]
={(x²+1-x²+x)/[x(x-3)]}×[x/(x+1)]
={(x+1)/[x(x-3)]}×[x/(x+1)]
=1/(x-3)
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