化简求值2x²y-[x²y+2(x²y-xy)]+y,其中x²=2x+1,y=2011/2012

化简求值2x²y-[x²y+2(x²y-xy)]+y,其中x²=2x+1,y=2011/2012
化简求值2x²y-[x²y+2(x²y-xy)]+y,其中x²=2x+1,y=2011/2012

化简求值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
=2x²y-(3x²y-2xy)+y
=2x²y-3x²y+2xy+y
=-x²y+2xy+y
=-y(x²-2x-1)
=-y[x²-(2x+1)]
=-(2011/2012)(x²-x²)
=0

原式=2x²y-[x²y+2xy（x-1）]+y=2x²y-[xy（x+2x-2）]+y=-x²y+2xy+y=y（1+2x-x²）=0

2x²y-[x²y+2(x²y-xy)]+y
=2x²y-[x²y+2xy(x-1)]+y
=2x²y-[x²y+2xy*2x]+y
=-3x²y+y

后面不会了