2.{[(4x+3)/(x-2)]^2}-{[(4x-3)/(x-2)]^2}=
3.a^2+amb-an-nc+anc-mnb=
咩是第一4分位數?
咩是第三4分位數?
咩是第65百分位數?
更新1:
請高手幫幫我 要詳細d~thx!
更新2:
第3題其實無錯-口- 所以我都唔明 唔識做...
因式分解同少少數學....問題 請各位幫手
2008-07-17 4:50 am
1.因式分解 a^2+2ab+2a+(b+1)^2=
2.{[(4x+3)/(x-2)]^2}-{[(4x-3)/(x-2)]^2}=
3.a^2+amb-an-nc+anc-mnb=
咩是第一4分位數?
咩是第三4分位數?
咩是第65百分位數?
2.{[(4x+3)/(x-2)]^2}-{[(4x-3)/(x-2)]^2}=
3.a^2+amb-an-nc+anc-mnb=
咩是第一4分位數?
咩是第三4分位數?
咩是第65百分位數?
回答 (2)
2008-07-17 6:55 am
✔ 最佳答案
1.a^2 +2ab +2a + (b+1)^2 = a^2 + 2ab + 2a + b^2 + 2b + 1
= a^2 + 2ab + b^2 + 2a + 2b + 1
= (a + b)^2 + 2(a + b) + 1
= [(a +b) + 1 ] ^2 = (a + b + 1)^2 .
2.
[(4x +3)/(x-2)]^2 - [(4x - 3)/(x-2)]^2
= [1/(x-2)]^2 [(4x +3)^2 - (4x -3)^2]
=[1/(x-2)]^2 [(4x +3 + 4x -3)(4x +3 - 4x +3)]
= [1/(x-2)]^2 (8x)(6) = 48x/(x-2)^2.
3.
Is it really -nc and not -n^2c?
4. They are first quartile, third quartile and 65 percentile.
Third quartile - first quartile = interquartile range.
2008-07-17 6:39 am
1. (a)^2 +2a(b+1) + (b+1)^2
= [a+(b+1)]^2
2. 通分母 (其實不用通, 因為是一樣) then, 分子是 in a^2-b^2 format, 所以變成 (a+b) (a-b), 即: [4x+3+(4x-3)][4x+3-(4x-3)]
= 48x
3. 有無打錯題目? 我factorize 不到
= [a+(b+1)]^2
2. 通分母 (其實不用通, 因為是一樣) then, 分子是 in a^2-b^2 format, 所以變成 (a+b) (a-b), 即: [4x+3+(4x-3)][4x+3-(4x-3)]
= 48x
3. 有無打錯題目? 我factorize 不到
收錄日期: 2021-04-25 22:35:28
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