[(x-1/2y)²-1]-[x-1/2y-1]还能化简吗[(x-1/2y)²-1]-[x-1/2y-1]²是这个式子

[(x-1/2y)²-1]-[x-1/2y-1]还能化简吗
[(x-1/2y)²-1]-[x-1/2y-1]²是这个式子

首先,题目的后项是不是少了个括号?是[(x-1)/2y-1]?是1/(2y).还是 (1/2)y
[(x-1/2y)²-1]-[(x-1)/2y-1]
=(x-1)²/4y²-1-[(x-1)/2y-1]
=[(x-1)²-4y²]/4y²-[(x-1)/2y-1]
=(x-1+2y)(x-1-2y)/4y²-[(x-1)/2y-1]
=(x-1+2y)(x-1-2y)/4y²-(x-1-2y)*2y/4y²
=(x-1+2y)[(x-1-2y)/4y²-2y/4y²]
=(x-1+2y)(x-1-4y)/4y²

=X²-XY+1/4Y²-1-X+1/2y+1
=X²-XY+1/4Y²-X+1/2y

[(x-1/2y)²-1]-[x-1/2y-1]²
=[(x-1/2y)²-1]-[(x-1/2y)²-1²]
=0[(x-1/2y)²-1]-[x-1/2y-1]²
=[(x-1/2y)²-1]-[(x-1/2y)²-1²] 这一步骤，没有什么公式啊
=0<...

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[(x-1/2y)²-1]-[x-1/2y-1]²
=[(x-1/2y)²-1]-[(x-1/2y)²-1²]
=0

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