想問大家一條數,,(頁 1) - 吹水版 - 娛樂滿紛 26FUN

SKII999 發表於 2008-1-17 07:26 PM

想問大家一條數,,

ab+bc-ac=0
      a-c=1017u.A-gZcizE
      求b M;KHyZ1wp/p

      要STEPS,,,THX

托迪仔T 發表於 2008-1-17 07:42 PM

a=101, b&c =0 :D

SKII999 發表於 2008-1-17 07:52 PM

我都知,,但唔知點得出個ans.,,anyway,,THX

ch.chak 發表於 2008-1-17 09:38 PM

a=101-c
ab+bc-ac=07P"c*w:M sGL
b(101-c)+bc-c(101-c)=0t5o2m4ty*B5i
101c-bc+bc-101c+(c*c)=0 P*o8l+E6f.M_$BDl
c*c=0
c=0

a=101-07h.~t,yj@/p
a=101
9~.s(v)eHzA{9H
ab+bc-ac=0
a=101
c=0EZ,KDs {v
101b + 0*b - 101*0=0)J4Qv^%L_|/[
101b=03x\o+[3}Et7H`
b=0

sakura310 發表於 2008-1-17 09:58 PM

[quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:
a=101-c
ab+bc-ac=0 v9r+mY Y`7u)]
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0 F&NAx/`p
c*c=0
c=0+vYF(M-og pJ5NXj

a=101-0%[3Ak K&AHvu]V
a=101
Tj6U {z-E3^:D:Z Op
ab+bc-ac=0(b&`n ?/j A
a=1010wrs*ID,d9UFEX
c=0
101b + 0*b - 101*0=0
101b=0
b=0 [/quote] JvW`vO`J K \

對不起Gtc!DrB
請容許我指出錯誤
E2EzSa iv
在一開頭

[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)t1k `R[H*`7K q
ab+bc-ac=0sC C3G8Y,@
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0_0}"nEmC!?
[color=Red]101c[/color]-bc+bc-101c+(c*c)=0
c*c=0/j Wc8uI
c=0Y}'{I4Ge

基於以上紅字的地方,其餘的地方我認為沒必要看了

ch.chak 發表於 2008-1-17 10:53 PM

[quote]Originally posted by [i]sakura310[/i] at 2008-1-17 09:58 PM:-tk N)tNr%z7o

對不起
請容許我指出錯誤j@ C;B+F8I#G2bM

在... [/quote]
係bo aV3Y&F7Wa Eni`
冇留意錯左tim

呀俊發表於 2008-1-17 11:00 PM

[quote]Originally posted by [i]托迪仔T[/i] at 2008-1-17 07:42 PM:Pc"vwh5Z~p;[
a=101, b&c =0 :D [/quote]P_bT,e$N1b'A k
ab+bc-ac=0:?f'p6?J9l
a-c=101 Ek \k(w)Q
求b
咁玩法...如果c係-101...a同b=0 都得架wo:D

playbr2 發表於 2008-1-17 11:58 PM

[quote]Originally posted by [i]sakura310[/i] at 09:58 PM:

zu.Cr3]B/i|
對不起
請容許我指出錯誤

在... [/quote]

咁你又吾去計;);):P:P

sakura310 發表於 2008-1-18 12:08 AM

[quote]Originally posted by [i]playbr2[/i] at 2008-1-17 11:58 PM:
I?z]8@
6v8C vC$]g
咁你又吾去計;);):P:P [/quote]
m X*kE3_a8MX
playplay你串我?:mad::noway::fight::angry:(表情符號好好玩的說:D)
(s6pyHj-xhj
咁你呢~你又唔去計?:dev::dev:

等了這麼久也沒人貼答案出來Sz&{/r[\]
pa L2TT
那麼讓小生物來獻醜了:redface::redface::redface:j3j`.B L&S J;aW

手寫關係,所以上傳圖片先了(懶得去打)Vv#I7@+o6L e+C6|
!a3?@wuEs h
由於題目沒有說到a,b和c是甚麼數,我暫且假設係整數黎做(光是實數的話...太沒意思了...) DC+jm+gM5C3Q
Kmadv"B+l!D
如果是實數的話,那麼請看到左邊2/3就可以了,中間和右邊是整數的解集(次序:左>中>右)

用的不是甚麼古怪的方法,應該好易睇得明sK9C1d"c9lH6zi_
8j,H\i#~4`
這算一種對付三個未知數,但只有兩式的求整數解方法
(j)NZ6P/k0pD(_(R
有不明的話可以提出

不過我要訓醒頭覺先回覆到,好累的說(脫力中...)

[[i] Last edited by sakura310 on 2008-1-18 at 12:12 AM [/i]]

playbr2 發表於 2008-1-18 12:21 AM

[quote]Originally posted by [i]sakura310[/i] at  12:08 AM:
{2s)Bo't6p
lAdL#K;g
playplay你串我?:mad::noway::fig... [/quote]
y;Ig!X7o)G*t
乜咁快手ㄚㄚㄚㄚ:D:D:D:D

個答案可以咁.....Qy{5peWA\$h
.HAEO1L/K
1.ab+bc-ac=0'A4?:o [7d
a-c=101T[ Oll4q({(v
b=ac/(a+c)=(101c+c^2)/(101+2c):c'[3V@,V]l;]3{
設c=101nsh:c+wb3wt
101n代入cx;q,Z$}:\pq

[101^2*n+(101n)^2]/(101+202n)
=[(101^2)n(n+1)]/[(101)(2n+1)]
=[101n(n+1)]/(2n+1)&[.{:I4_v0Q IZ.q;P

因為a、b、c是正整數
所以n是正整數c2_OJ `4c{ a
所以[101n(n+1)]/(2n+1)是整數"VIi d5T
2n+1=101(剩下的2n+1=n或2n+1=n+1，n沒有整數解)
n=50^(pTfi,_^ kSc
a=5151  <--a-c=101Q7N4qd^0R
b=2550  <--[101n(n+1)]/(2n+1)F A[(\o1auA5]
c=5050  <--c=101n6O-wIbR`
答案:b=2550

sakura310 發表於 2008-1-18 12:27 AM

Filename: DSC01158.JPG
Datetime | Downloaded by | CountsB$Sj P9sO!d@
2008-01-18 00:25:24 | 阿感 | 2
2008-01-18 00:24:20 | playbr2 | 2
2008-01-18 00:22:02 | 但丁 | 2
,KX4kX'j
點解大家都down兩次? XD

playbr2 發表於 2008-1-18 12:30 AM

[quote]Originally posted by [i]sakura310[/i] at 12:27 AM:
Filename: DSC01158.JPG
Datetime | Downloaded by | Counts{l3jFj B:K6z([
2008-01-18 00:25:24 | 阿感 | 2
2008-01-18 00:24:20 | playbr2 | 2+o:^@"G1K
2008-01-18 00:22:02 | 但丁 | 2J_/E2O;o4q

點解大家都down兩次? XD [/quote]

我第一次開睇左2秒就close 左:P:P
之後DL (就变成第二次)

playbr2 發表於 2008-1-18 12:34 AM

[quote]Originally posted by [i]sakura310[/i] at 12:27 AM:O)aJ3zH}BN
Filename: DSC01158.JPG&]p.KD_rXn`*A
Datetime | Downloaded by | Counts
2008-01-18 00:25:24 | 阿感 | 2e5OoL6|Y3?
2008-01-18 00:24:20 | playbr2 | 2
2008-01-18 00:22:02 | 但丁 | 2

點解大家都down兩次? XD [/quote]
?*\c8gt;CbN|J
快d捉我錯處:D:D:D
話吾定我都打錯 + - 符號
係電腦打答案...的確幾煩...又哂時間...:giveup:

sakura310 發表於 2008-1-18 06:47 AM

捉錯處就唔敢當M]qp C
不過,c=101n哥度(解除n係正整數的限制,只是整數)#]FzH(k"wz!Fk U
就一下子限制左c一定要係101的倍數(好像看懸疑片,突然劇情說出一些事前沒提及過的線索,不爽~)
知道{101,0,0}和{5151,2550,5050}係答案的話,當然沒問題
但原先是不知答案,所以唔知有無失根

有時見到題目都感到有點不爽
題目所求的答案是整數,正整數,有理數,實數等等應該說清楚
有些題目,整數解和正整數解分別好大的說,要找出整數解集有時可不是件易事= =

cutec 發表於 2008-1-18 08:26 AM

冇答案ga.... |Z3\4e5k$](w_w
.x9Fdq"I6?:aM#M
2條equation solve 唔到3個unknown ga....

Azuma 發表於 2008-1-18 08:56 AM

[quote]Originally posted by [i]cutec[/i] at 2008-1-18 08:26 AM:'JPc,\s-sc] oRV
冇答案ga....&J WO`:V|
(b)O%ei [f&at;w7S`C
2條equation solve 唔到3個unknown ga.... [/quote]
x9u#]3[%}] eN
agree

2 equations 2 unknows?)z Y/b@5pwJaV

if 3 unknows there should be 3 equations

make it simple N*q(d0C0V^ is^:L0m

say `5Qqm|-g4B#uS

a + b +c = 3'_#qZM#d

a+b = 2
-m,r j!l T
then c is 1
VZ}k%QH+\n N.G
a = 2 - b
*Pex-O7|{cf
so 0 = 0 BtC,yQ&IG5x
!\U9u0qE
:giveup:

sakura310 發表於 2008-1-18 10:22 AM

[quote]Originally posted by [i]cutec[/i] at 2008-1-18 08:26 AM:
冇答案ga....

2條equation solve 唔到3個unknown ga.... [/quote]V%u$_N4k-k/f\i

此言差矣~
-U/gVX)X
2條equation solve 唔到3個unknown ga3z.\ p2y(`/[#l:U
還欠一個條件,就是3條equation都是linear independent(線性無關)的K|(oV[8\ {\!m^

簡單點說,多元一次方程就是線性方程的一個例子
在這情況下,當然是想求n個未知數,至少要用n條線性無關的方程"jhs#ot(} CW
:V;h+k NLX#~
如果不是線性的話,有機會1條方程解3個未知數的(這不是每次都可以的啦~)F yY7P2Yb(mz
例如 x^2-2(x+8y)+17+4y^2+z^0.5=0 就只有唯一整數解
有興趣就解解看:sleep:e,m`Z3VRKX],B/a
不懂的話就問playplay,佢會知答案:dev:

playbr2 發表於 2008-1-18 02:47 PM

[quote]Originally posted by [i]sakura310[/i] at 10:22 AM:

[vh#J(~$]l:t
不懂的話就問playplay,佢會知答案[/quote]m:o0{~+e#i
f`UU+g
:cry::giveup::giveup::giveup::giveup:

playbr2 發表於 2008-1-18 02:57 PM

[quote]Originally posted by [i]sakura310[/i] at 06:47 AM:a5Yq7O&~m0H
捉錯處就唔敢當
不過,c=101n哥度(解... [/quote]@(D_EbkDc(g
i W M6z^%jM+kp
都係睇餸食飯姐 , 無理由因為題目冇同你假設就下下大包圍架
反而適當地調節一下 , 咁會理性好多2~"V.d a3i
就好似隻game 由12粒珠加到13粒珠咁
當''堂''好玩哂:P:P:P
2Z5~{FC0k/[y?v
p.s. 求''堂''字正寫:agree::giveup:

QuiGonJinn 發表於 2008-1-19 03:49 AM

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娛樂滿紛 26FUN's Archiver

想問大家一條數,,