想問大家一條數,,
ab+bc-ac=0*n+^3| T+XA-~a-c=101
求b
要STEPS,,,THX a=101, b&c =0 :D 我都知,,但唔知點得出個ans.,,anyway,,THX a=101-cO0pE3N,ja
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0s7L;BZw _b
101c-bc+bc-101c+(c*c)=0S0jxw4_6z7Z"p%Y
c*c=0
c=0
a=101-0
a=101 lr{xJUW
ab+bc-ac=0
a=101
c=0VD*^1vp)}/Rm|
101b + 0*b - 101*0=06m0j NDX5C$Y1oL H(V
101b=0
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:(V'_"~$E_G+h{
a=101-c
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0$n0L?Rt
101c-bc+bc-101c+(c*c)=0
c*c=0
c=0
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a=101-0
a=101
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ab+bc-ac=0
a=101tE.\&v:BZ1F
c=0)o8g.eT`&|B+xDx5X
101b + 0*b - 101*0=0"X \I!` ^_ o
101b=0
b=0 [/quote]
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對不起l?h+_&m8HN
請容許我指出錯誤o5Y;Gv ug$A;{,E~4I
在一開頭
[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)m9J;r9m(|ajGs
ab+bc-ac=0
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=06B [.LLX;q)G5^%z[K8z
[color=Red]101c[/color]-bc+bc-101c+(c*c)=0^kG)yX)m`}
c*c=0g!['R5i e A"O
c=0'X'h-z H _"c/b
C"R+G F5V#M]3qgA
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:
對不起.@u!mw!q.}F z/RS"H
請容許我指出錯誤
在... [/quote]
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冇留意錯左tim [quote]Originally posted by [i]托迪仔T[/i] at 2008-1-17 07:42 PM:
a=101, b&c =0 :D [/quote]!oX)n&hg
ab+bc-ac=0 w(Ne"F5p/Z1O c^
a-c=101 ]+}2q&tO
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:
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對不起
請容許我指出錯誤4t8b9q#b NH`"P
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在... [/quote]%Rx$Y5R`6D%L&hn?+[
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咁你又吾去計;);):P:P [quote]Originally posted by [i]playbr2[/i] at 2008-1-17 11:58 PM:/?,C H!Ry@q
咁你又吾去計;);):P:P [/quote]
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playplay你串我?:mad::noway::fight::angry:(表情符號好好玩的說:D)mtZ hwN
Z6d%a&x.vB ~|
咁你呢~你又唔去計?:dev::dev:
等了這麼久也沒人貼答案出來:Nj\u+yN
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那麼讓小生物來獻醜了:redface::redface::redface:?Y9R,TFyo [[G
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手寫關係,所以上傳圖片先了(懶得去打)[ m&hI*L9l'GS
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由於題目沒有說到a,b和c是甚麼數,我暫且假設係整數黎做(光是實數的話...太沒意思了...)
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如果是實數的話,那麼請看到左邊2/3就可以了,中間和右邊是整數的解集(次序:左>中>右)
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用的不是甚麼古怪的方法,應該好易睇得明
這算一種對付三個未知數,但只有兩式的求整數解方法,J0N7?6}v&W
有不明的話可以提出
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不過我要訓醒頭覺先回覆到,好累的說(脫力中...)
[[i] Last edited by sakura310 on 2008-1-18 at 12:12 AM [/i]] [quote]Originally posted by [i]sakura310[/i] at 12:08 AM:
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playplay你串我?:mad::noway::fig... [/quote]
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乜咁快手ㄚㄚㄚㄚ:D:D:D:D@ ]V*Oe1K\i5v
個答案可以咁.....,YGL(P$w3ZK%n#R'f
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1.ab+bc-ac=0O0|k1b vrh
a-c=101
b=ac/(a+c)=(101c+c^2)/(101+2c)wa0f_$_(jv%b
設c=101n?.m3b,e&D%v
101n代入c[I3J _5f H
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[101^2*n+(101n)^2]/(101+202n)
=[(101^2)n(n+1)]/[(101)(2n+1)]{#pn'SE/ob)WW2j3Vq
=[101n(n+1)]/(2n+1)1VSE{@k!y3P/r2p
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因為a、b、c是正整數8Gl:s,n"S6?u)y
所以n是正整數'r1J%I7}"x1]wYG
所以[101n(n+1)]/(2n+1)是整數
2n+1=101(剩下的2n+1=n或2n+1=n+1,n沒有整數解)ikz)qQ$J_ xH
n=50
a=5151 <--a-c=101
b=2550 <--[101n(n+1)]/(2n+1)b-y$Q'a})N%q r8w&L,Z
c=5050 <--c=101n
答案:b=2550 Filename: DSC01158.JPG
Datetime | Downloaded by | Counts8t];@ E h
2008-01-18 00:25:24 | 阿感 | 2
2008-01-18 00:24:20 | playbr2 | 2
2008-01-18 00:22:02 | 但丁 | 2e}5at&hg
O%p$s]O6d Q?R2a$B
點解大家都down兩次? XD [quote]Originally posted by [i]sakura310[/i] at 12:27 AM:xW WO,Z
Filename: DSC01158.JPG
Datetime | Downloaded by | Counts
2008-01-18 00:25:24 | 阿感 | 2
2008-01-18 00:24:20 | playbr2 | 22sMWr0^ss3s
2008-01-18 00:22:02 | 但丁 | 2
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點解大家都down兩次? XD [/quote]
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我第一次開睇左2秒就close 左:P:P
之後DL (就变成第二次) [quote]Originally posted by [i]sakura310[/i] at 12:27 AM: