想問大家一條數,,
ab+bc-ac=0a-c=101B.Cc)J(G} n,l
求bP m\n6W v
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
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0g)q1]#g @@n^
101c-bc+bc-101c+(c*c)=0UG;xEu
c*c=0
c=0
a=101-07ty n!O K4lJC
a=101R%lF9`.~ ?Q6I%s$OF`
ab+bc-ac=0C j} nP(NItx
a=101KLMD:J&QX] ?
c=0,]IY;jhs
101b + 0*b - 101*0=0*j)?'K3P*Z%x]8S
101b=0.^p{5H[+Z7M
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:
a=101-c
ab+bc-ac=0m)Hf_3e7yD)KE
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0f3`7P9rT F0w5LW
c*c=0
c=0p}d7U8J$}+\p
%r]V#lCT#IM&P
a=101-0#C8X(G%E HSnn
a=101 VD.c(p P n
M:~I+q-qG$Jj3Dm
ab+bc-ac=0
a=101Z.o%g'cAe
c=0&v+e'Uw U.b5Lz c
101b + 0*b - 101*0=0I-tosbR
101b=0
b=0 [/quote])j.g@6coj.? x'@&F
_PxN Bb He
對不起
請容許我指出錯誤
在一開頭
g#Eza)qn$w
[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)
ab+bc-ac=0#I&A!]jxZNmM
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0KW z|8k&M+z
[color=Red]101c[/color]-bc+bc-101c+(c*c)=0@,V"{3kr
c*c=0
c=0? c9IS0f
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:2]4uw;iO+mu @6k
對不起d j H,cpTD
請容許我指出錯誤
在... [/quote]g _+E'_c H"L"MS
係bo
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]t _)~r n!SG
ab+bc-ac=0U4I*T,mb ]-N~ _
a-c=101
求b+N~,MyS&Q\mQ
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:+^%O6T\4H$S m\3i