[x+2y-1][x-2y+1]-[x-2y-1]的平方
问题描述：
[x+2y-1][x-2y+1]-[x-2y-1]的平方
答
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
答
[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
答
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