1) y^2=x^2-8x+16
2)d/dx(3x^2-1/x)
更新1:
以上兩題是differential equations, 要用differential 的方法計,謝謝.
數學題, 請幫幫手
回答 (3)
2009-11-10 7:04 am
✔ 最佳答案
1) y^2 = x^2 - 8x + 16y^2 = (x - 4)^2
y^2 - (x - 4)^2 = 0
(y - x + 4)(y + x - 4) = 0
Therefore y = x - 4 or y = 4 - x
And dy/dx = 1 or dy/dx = -1
2) d/dx(3x^2 - 1/x)
= 6x + 1/x^2
2009-11-10 6:25 am
y^2=x^2-8x+16
=x^2-2(x)(4)+4^2
y^2=(x-4)^2
y=(x-4)
第二題我唔明u想ask咩bor,可否解釋一下?
=x^2-2(x)(4)+4^2
y^2=(x-4)^2
y=(x-4)
第二題我唔明u想ask咩bor,可否解釋一下?
參考: me
2009-11-10 4:18 am
y^2=x^2-8x+16
=x^2-2(x)(4)+4^2
=x^2-4^2
y^2=(x+4)^2
y=(x+4)
=x^2-2(x)(4)+4^2
=x^2-4^2
y^2=(x+4)^2
y=(x+4)
收錄日期: 2021-04-23 23:22:29
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