2x²y-[x²y+2（x²y-xy)]+y 其中x²=2x+1 y=2011/2012
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2x²y-[x²y+2（x²y-xy)]+y其中x²=2x+1y=2011/20122x²y-[x²y+2（x²y-xy)]+y其
2x²y-[x²y+2（x²y-xy)]+y 其中x²=2x+1 y=2011/2012
2x²y-[x²y+2（x²y-xy)]+y 其中x²=2x+1 y=2011/2012
2x²y-[x²y+2（x²y-xy)]+y 其中x²=2x+1 y=2011/2012
2x²y-[x²y+2（x²y-xy)]+y
=2x²y-x²y-2x²y+2xy+y
=-x²y+2xy+y
=-y[x²-(2x+1）]
∵x²=2x+1
∴2x²y-[x²y+2（x²y-xy)]+y=0