[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],

angushja 1年前 已收到1个回答 举报

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[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)]
=[1/(x-1)²-1/(x+1)²]÷[1/(x+1)+1/(x-1)]
=[1/(x+1)+1/(x-1)][1/(x-1)+1/(x+1)]÷[1/(x+1)+1/(x-1)]
=1/(x-1)-1/(x+1)
=(x+1)/(x+1)(x-1)-(x-1)/(x-1)(x+1)
=(x+1-x+1)/(x-1)(x+1)
=2/(x+1)(x-1)

提示:用平方差公式,不要通分

1年前

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