問功課時間- -

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[[i] 本帖最後由 話說從前 於 2008-12-29 07:23 PM 編輯 [/i]]

(x+2)^2-4(x^2-4x-12)

= (x+2)^2 - 4(x-6)(x+2)

= (x+2)^2 - (4x-24)(x+2)

= (x+2)(x+2-4x+24)

= (x+2)(26-3x)


唔知係唔係呢x_x" 最驚尼d囧.

[quote][b]原帖由 [i]`乜料[/i] 於 2008-12-29 06:11 PM 發表。 [/b][url=http://forum.eyankit.com/viewthread.php?tid=32973&page=1#pid320672][img]http://forum.eyankit.com/images/common/back.gif[/img][/url]
(x+2)^2-4(x^2-4x-12)

= (x+2)^2 - 4(x-6)(x+2)

= (x+2)^2 - (4x-24)(x+2)

= (x+2)(x+2-4x+24)

= (x+2)(26-3x)


唔知係唔係呢x_x" 最驚尼d囧. [/quote]
恭喜你,你得獎了:em28:

3.
角EAB = 角DBC ( given )
角EAB + 角AEB = 角EBD + 角DBC ( ext. 角 of [u]/\[/u]
角AEB = 角EBD
          = 90度
所以三角形ABE是直角三角形 <<自己轉英文啦:biggrin2:

唔知岩唔岩- -'

[quote][b]原帖由 [i]`乜料[/i] 於 2008-12-29 06:27 PM 發表。 [/b][url=http://forum.eyankit.com/viewthread.php?tid=32973&page=1#pid320695][img]http://forum.eyankit.com/images/common/back.gif[/img][/url]
3.
角EAB = 角DBC ( given )
角EAB + 角AEB = 角EBD + 角DBC ( ext. 角 of /\
角AEB = 角EBD
          = 90度
所以三角形ABE是直角三角形 <<自己轉英文啦:biggrin2:

唔知岩唔岩- -' [/quote]
很好- -'
thz

回覆 #4 `乜料 的帖子

乖b好叻叻牙a_a

[quote][b]原帖由 [i]w1ngYaN!`[/i] 於 2008-12-29 07:41 PM 發表。&nbsp;[/b][url=http://forum.eyankit.com/viewthread.php?tid=32973&page=1#pid320774][img]http://forum.eyankit.com/images/common/back.gif[/img][/url]
乖b好叻叻牙a_a [/quote]
:biggrin2:其他唔做係因為唔識

[quote][b]原帖由 [i]`乜料[/i] 於 2008-12-29 08:42 PM 發表。 [/b][url=http://forum.eyankit.com/viewthread.php?tid=32973&page=1#pid320832][img]http://forum.eyankit.com/images/common/back.gif[/img][/url]

:biggrin2:其他唔做係因為唔識 [/quote]
0TZ'
感謝你參與=]

(x+2)^2-4(x^2-4x-12)

= (x+2)^2 - 4(x-6)(x+2)

= (x+2)^2 - (4x-24)(x+2)

= (x+2)(x+2-4x+24)

= (x+2)(26-3x)




我想知系第幾題牙∼！！

[quote][b]原帖由 [i]快樂的人[/i] 於 2008-12-29 09:36 PM 發表。&nbsp;[/b][url=http://forum.eyankit.com/viewthread.php?tid=32973&page=1#pid320883][img]http://forum.eyankit.com/images/common/back.gif[/img][/url]
(x+2)^2-4(x^2-4x-12)

= (x+2)^2 - 4(x-6)(x+2)

= (x+2)^2 - (4x-24)(x+2)

= (x+2)(x+2-4x+24)

= (x+2)(26-3x)




我想知系第幾題牙∼！！ [/quote]
自己思索一下吧:clown:

一日一推身體好:em33:

再推:em33:

9.a
BE=ED(given)
EC=EC(common side)
<ECB=180-<ECD
<ECB=90
so,triangleECB=triangleECD(R.H.S)
As,triangleECB=triangleECD(R.H.S)
So,BC+CD=BD
As,BC=CD
So,BC=BD-CD
BC=1/2BD
唔知可唔可以咁証明- -"

b.As,BC=CD
BC=BD/2
So,CD=BD/2
AD=AE+ED
AE=ED
ED=AD/2
<EDC=<ADB(common angle)
CD/EB=BD/2/AD/2
=BD/AD
So,triangleABD~triangleECD(ratio of 2 side inc.<)

7a
GQSH is a straight line.(given)
<GQP+<PQS=180(adj. <s on st. line)
<PQS=90
<PQS+RSQ=180
So,AB//CD(int.<s supp.)

b.
As,AB//CD
3x-12=5x/2(alt.<s,AB//CD)
6x-24=5x
x=24

[[i] 本帖最後由 `乜囧. 於 2009-1-4 01:01 AM 編輯 [/i]]