题目内容
填空:(______)(1+a)=a2-1 x2+x+______=(x+
)2
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∵a2-1=(a+1)(a-1),
∴a-1)(1+a)=a2-1;
∵(x+
)2=x2+x+
,
∴x2+x+
=(x+
)2.
∴a-1)(1+a)=a2-1;
∵(x+
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| 2 |
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| 4 |
∴x2+x+
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| 4 |
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| 2 |
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