计算:(1)x+2yx2−y2+yy2−x2−2xx2−y2-查字典问答网
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来自蒋易的问题

  计算:(1)x+2yx2−y2+yy2−x2−2xx2−y2;(2)4−(15+2)0+(−2)3÷4−1;(3)x2+2x+1x+2÷x2−1x−1−1x+2;(4)(xx−1−2xx2−1)÷xx−1.

  计算:

  (1)x+2yx2−y2+yy2−x2−2xx2−y2;

  (2)

  4−(1

  5+2)0+(−2)3÷4−1;

  (3)x2+2x+1x+2÷x2−1x−1−1x+2;

  (4)(xx−1−2xx2−1)÷xx−1.

1回答
2020-09-09 02:26
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钱惟贤

  (1)原式=x+2yx2−y2-yx2−y2-2xx2−y2=x+2y−y−2xx2−y2=−x+y(x+y)(x−y)=-1x+y;(2)原式=2-1-8÷14=2-1-32=-31;(3)原式=x+1x+2-1x+2=xx+2;(4)原式=1-2x(x+1)(x−1)•x−1x=1-2x+1=x+1−2x+1=x−1x+1....

2020-09-09 02:27:52

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