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+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