想問大家一條數,,
ab+bc-ac=0!T4N1L bT!la-c=101
求bl$zB1Xk
8k|b4c%{eCf
要STEPS,,,THX a=101, b&c =0 :D 我都知,,但唔知點得出個ans.,,anyway,,THX a=101-c:Udq La9W6h
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0C4]p.Z&qt6|
101c-bc+bc-101c+(c*c)=0
c*c=0q1e3Uk9x @u
c=0fG3N+j}%q u
a=101-0
a=101
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ab+bc-ac=0 M1v|| m-|9S
a=101
c=0
101b + 0*b - 101*0=0!h!Ynm2D9cL3^
101b=0W-BP&?/d;z
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:
a=101-c
ab+bc-ac=0,I/k(rsO(HM
b(101-c)+bc-c(101-c)=0$e}6M2sNJ4N d
101c-bc+bc-101c+(c*c)=02e5c ]8ge.BHuQ
c*c=0'L8|:xFf)L4T{
c=0&qN#n/F9I$c
f4@,K.k-M+} v6v
a=101-0*Z#Y2?a*T)N"I
a=101 j0N/? rT2GJ6Xu
ab+bc-ac=0
a=101
c=09f8V$}3f%oh.[9`0}V+dZ.q
101b + 0*b - 101*0=0
101b=0
b=0 [/quote]
對不起
請容許我指出錯誤
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在一開頭
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[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)
ab+bc-ac=0|4kc]J~;R
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0"Q_ X uu `,u
[color=Red]101c[/color]-bc+bc-101c+(c*c)=0 QeyHn/a!o @-@)a4f+Mm
c*c=0
c=0.x ^:r @;K5d
基於以上紅字的地方,其餘的地方我認為沒必要看了 [quote]Originally posted by [i]sakura310[/i] at 2008-1-17 09:58 PM:[l;IBVR
7j3WET M]n3B$^
對不起8^Zfc1KP
請容許我指出錯誤
在... [/quote]4{2MA*O"KR3e
係bo7dKX$etE
冇留意錯左tim [quote]Originally posted by [i]托迪仔T[/i] at 2008-1-17 07:42 PM:2IZ/O/d SO
a=101, b&c =0 :D [/quote]
ab+bc-ac=0
a-c=101
求b
咁玩法...如果c係-101...a同b=0 都得架wo:D [quote]Originally posted by [i]sakura310[/i] at 09:58 PM:
對不起$hCit5J
請容許我指出錯誤6u"F;^~3wT'vN
在... [/quote]
[nh'y)\ [0o Z+t8G+t$l,u
咁你又吾去計;);):P:P [quote]Originally posted by [i]playbr2[/i] at 2008-1-17 11:58 PM:
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咁你又吾去計;);):P:P [/quote]
playplay你串我?:mad::noway::fight::angry:(表情符號好好玩的說:D)
咁你呢~你又唔去計?:dev::dev:
等了這麼久也沒人貼答案出來fP}J6DQ n9Y
那麼讓小生物來獻醜了:redface::redface::redface:
%Z_&A`@&rnct
手寫關係,所以上傳圖片先了(懶得去打)
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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...)
L Ad G/L$W;zQyc'b
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)
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用的不是甚麼古怪的方法,應該好易睇得明
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這算一種對付三個未知數,但只有兩式的求整數解方法"]@/I0\ [5jB%H/?
有不明的話可以提出5]oA{y i#W|L4w
不過我要訓醒頭覺先回覆到,好累的說(脫力中...)
m#O^ WU+F
[[i] Last edited by sakura310 on 2008-1-18 at 12:12 AM [/i]] [quote]Originally posted by [i]sakura310[/i] at 12:08 AM:#J ]wmPk
$o0kCSSN;t/NM c
Or5{^UV
playplay你串我?:mad::noway::fig... [/quote]
乜咁快手ㄚㄚㄚㄚ:D:D:D:D
Vxgn-WI+{
個答案可以咁.....L zhm6l'j)`
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1.ab+bc-ac=0$V0}5c|g;a`E/@
a-c=1012v*W6HqG {f
b=ac/(a+c)=(101c+c^2)/(101+2c)%nV-x&? G)Iev8C:k
設c=101n
101n代入c
[101^2*n+(101n)^2]/(101+202n)
=[(101^2)n(n+1)]/[(101)(2n+1)]5a.NJ2g~IGF|jp
=[101n(n+1)]/(2n+1)qe`MB:v~Y
因為a、b、c是正整數
所以n是正整數(o[r-Z-K
所以[101n(n+1)]/(2n+1)是整數
2n+1=101(剩下的2n+1=n或2n+1=n+1,n沒有整數解)-sa8o`7}O!| O{
n=50
a=5151 <--a-c=101r9h%sa F
b=2550 <--[101n(n+1)]/(2n+1)oN-dif!y ? x
c=5050 <--c=101n
答案:b=2550 Filename: DSC01158.JPG`2E?,k[
Datetime | Downloaded by | Counts
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'{mix1dn e8r'SJ
)C.~)E/^M,t6n&G!BjF
點解大家都down兩次? XD [quote]Originally posted by [i]sakura310[/i] at 12:27 AM:
Filename: DSC01158.JPG;l1S+F3s(f
Datetime | Downloaded by | Counts
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 | 22v/xv(])x^
點解大家都down兩次? XD [/quote]
^4?cR}:R
我第一次開睇左2秒就close 左:P:P