Question

初二数学：计算：[x-(1/x-y)]²-[x+(1/x-y)]²

Answer 1

[x-(1/x-y)]²-[x+(1/x-y)]²
={[x-(1/x-y)]+[x+(1/x-y)]}*{[x-(1/x-y)]-[x+(1/x-y)]}(平方差公式）
={x-(1/x-y)+x+(1/x-y)]}*{x-(1/x-y)-x-(1/x-y)}
=2x*[-(2/x-y)]
=-4x/(x-y)
希望采纳！

Answer 2

[x-(1/x-y)]²-[x+(1/x-y)]²
={[x-(1/x-y)]+[x+(1/x-y)]}*{[x-(1/x-y)]-[x+(1/x-y)]}(平方差公式）
={x-(1/x-y)+x+(1/x-y)]}*{x-(1/x-y)-x-(1/x-y)}（去括号）
=2x*[-(2/x-y)]（计算）
=4x/(x-y)（得出结果）

Answer 3

[x-(1/x-y)]²-[x+(1/x-y)]²={[x-(1/x-y)]+[x+(1/x-y)]}*{[x-(1/x-y)]-[x+(1/x-y)]}(平方差公式）={x-(1/x-y)+x+(1/x-y)]}*{x-(1/x-y)-x-(1/x-y)}=2x*[-(2/x-y)]=-4x/(x-y)

Answer 4

如果是化简的话最后是-4x/(x-y)