1 a .(2a - 3)(4a -11)+2
=(4a - 7)(2a - 5) 我計到
1b (2x^2 - 11)(4x^2 - 27)+2
ans=(4x^2 - 23)(2x^2 - 13)
我想問點計出來?
一條F.3 Maths,急
2012-10-14 10:02 pm
回答 (2)
2012-10-15 2:11 am
✔ 最佳答案
如果題目有from the result of a.就咁計出黎係冇分既b.(2x^2-11)(4x^2-27)+2
(2x^2-8-3)(4x^2-16-11)+2 為左砌番a個樣出黎
[2(x^2-4)-3][4(x^2-4)-11]+2
設a=x^2-4
(2a-3)(4a-11)+2
By result of a
=(4a-7)(2a-5)
=[4(x^2-4)-7][2(x^2-4)-5]
=(4x^2-23)(2x^2-13)
參考: 自己
2012-10-14 11:13 pm
(2x^2 - 11)(4x^2 - 27)+2
i.e. [(2x^2 - 11)(4x^2 - 27)]+2
(8x^4-54x^2-44x^2+972)+2
8(x^2)^2 -98x^2+299
cross method:
4x^2 -23
2x^2 -13
---------------
-46-52 = -98
=(4x^2 - 23)(2x^2 - 13)
p.s. 8x^4 = 8(x^2)^2
i.e. [(2x^2 - 11)(4x^2 - 27)]+2
(8x^4-54x^2-44x^2+972)+2
8(x^2)^2 -98x^2+299
cross method:
4x^2 -23
2x^2 -13
---------------
-46-52 = -98
=(4x^2 - 23)(2x^2 - 13)
p.s. 8x^4 = 8(x^2)^2
收錄日期: 2021-04-16 15:07:07
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