(a+1)^2+(a-2)^2
=？
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列式計算！
數學ｘ１（簡單） １０點
2007-07-31 5:40 am
回答 (3)
2007-07-31 7:02 am
(a + 1)^ 2 + (a - 2)^2
= (a^2 + 1^2 + 2*a*1) + (a^2 + (-2)^2 - 2*a*2)
= (a^2 + 2a + 1) + (a^2 - 4a + 4)
= 2a^2 - 2a + 5
= (a^2 + 1^2 + 2*a*1) + (a^2 + (-2)^2 - 2*a*2)
= (a^2 + 2a + 1) + (a^2 - 4a + 4)
= 2a^2 - 2a + 5
2007-07-31 5:48 am
(a+1)(a+1)+(a-2)(a-2)
a^2+2a+1+a^2-4a+4
2a^2-2a+5
a^2+2a+1+a^2-4a+4
2a^2-2a+5
2007-07-31 5:46 am
(a+1)^2+(a-2)^2
= (a^2+2a+2) + (a^2-4a+4)
= 2a^2 - 2a + 6
= 2(a^2-a+6)
=2(a-3)(a+2)
= (a^2+2a+2) + (a^2-4a+4)
= 2a^2 - 2a + 6
= 2(a^2-a+6)
=2(a-3)(a+2)
收錄日期: 2021-04-12 20:42:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070730000051KK05448