代數恆等式演變步驟 no.2
回答 (2)
2008-04-29 1:40 am
2x^2 + 3x + 1
=2x^2+2x+x+1 <---3x =>2x+x
=2x(x+1)+(x+1)
=(x+1)(2x+1)
=2x^2+2x+x+1 <---3x =>2x+x
=2x(x+1)+(x+1)
=(x+1)(2x+1)
2008-04-28 3:06 pm
2x^2+3x+1
=2x^2+3x+1^2 (將1加上2次)
=(2x+1)(x+1) (加上2次後,,就可以把2次項抽出)
=2x^2+3x+1^2 (將1加上2次)
=(2x+1)(x+1) (加上2次後,,就可以把2次項抽出)
參考: me
收錄日期: 2021-04-13 21:15:15
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