
计算【(a+b分之1)的2次方-(a-b分之1)的2次方】÷(a+b分之1-a-b分之1)
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{[1/(a+b)]²-[1/(a-b)]²}÷[1/(a+b)-1/(a-b)]
={[(a-b)²/(a²-b²)²]-[(a+b)²/(a²-b²)²]}÷{[(a-b)/(a²-b²)]-[(a+b)/(a²-b²)]}
=[-4ab/(a²-b²)²]÷[2b/(a²-b²)]
=[-4ab/(a²-b²)²]×[(a²-b²)/2b]
=-2a/(a²-b²)
x³y³-2x²y²+xy
=xy(x²y²-2xy+1)
=xy(xy-1)²
={[(a-b)²/(a²-b²)²]-[(a+b)²/(a²-b²)²]}÷{[(a-b)/(a²-b²)]-[(a+b)/(a²-b²)]}
=[-4ab/(a²-b²)²]÷[2b/(a²-b²)]
=[-4ab/(a²-b²)²]×[(a²-b²)/2b]
=-2a/(a²-b²)
x³y³-2x²y²+xy
=xy(x²y²-2xy+1)
=xy(xy-1)²
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四个选择木有这个答案哎
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{[1/(a+b)]²-[1/(a-b)]²}÷[1/(a+b)-1/(a-b)]
={[(a-b)²/(a²-b²)²]-[(a+b)²/(a²-b²)²]}÷{[(a-b)/(a²-b²)]-[(a+b)/(a²-b²)]}
=[-4ab/(a²-b²)²]÷[-2b/(a²-b²)]
=[-4ab/(a²-b²)²]×[-(a²-b²)/2b]
=2a/(a²-b²)
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