[x+2y-1][x-2y+1]-[x-2y-1]的平方
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[x+2y-1][x-2y+1]-[x-2y-1]的平方[x+2y-1][x-2y+1]-[x-2y-1]的平方[x+2y-1][x-2y+1]-[x-2y-1]的平方原式=[x+(2y-1)][x-
[x+2y-1][x-2y+1]-[x-2y-1]的平方
[x+2y-1][x-2y+1]-[x-2y-1]的平方
[x+2y-1][x-2y+1]-[x-2y-1]的平方
原式=[x+(2y-1)][x-(2y-1)]-[x-(2y+1)]²
=x²-(2y-1)²-x²+2x(2y+1)-(2y+1)²
=x²-4y²+4y-1-x²+4xy+2x-4y²-4y-1
=-8y²+4xy+2x-2
[x+2y-1][x-2y+1]-[x-2y-1]的平方
=[x+2y-1]{[x-2y+1]-[x-2y-1]}
=[x+2y-1]*2
=2x+4y-2
解 原式=[x+(2y-1)][x-(2y-1)]-[x-(2y+1)]²(平方差公式(a+b)*(a-b)=a²-b²)
=x²-(2y-1)²-x²+2x(2y+1)-(2y+1)²
=x²-4y²+4y-1-x²+4xy+2x-4y²-4y-1
=-8y²+4xy+2x-2
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