想問大家一條數,,
ab+bc-ac=0a-c=101
求b;jgcJM~/Q` ^
WF~*r3JB*o
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*QqJh)hN(h1B
b(101-c)+bc-c(101-c)=0#t4B7kcW
101c-bc+bc-101c+(c*c)=0#P+D\.Q z8|
c*c=0 m?E$\Sk
c=0]!K0Pze [a(iCfhx O
DA6N'X/K$yq
a=101-0
a=101;vfz!Jc
'?vmH+Y e
ab+bc-ac=0
a=101
c=0
101b + 0*b - 101*0=0![s.t)Z/RVw!s
101b=0
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:sB&b.wnJI_
a=101-c? f3G8}c i q/OU
ab+bc-ac=0?,l(CYoKQQ:G8K
b(101-c)+bc-c(101-c)=0js z0H;m Z W
101c-bc+bc-101c+(c*c)=0
c*c=0,`~~/?5Th'z2Jl(R
c=0,ze Cb9B+T q)x~
rx R \.b2Rm`
a=101-0
a=101
ab+bc-ac=0] R'tZ!H+W`1U)]3mHk
a=101
c=0
101b + 0*b - 101*0=0
101b=0p`_o*y0n VEr/e
b=0 [/quote]
對不起YPCc.AVk0j
請容許我指出錯誤
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在一開頭o9|0S#W$DN
a-IS"e uv8n
[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)
ab+bc-ac=0
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0
[color=Red]101c[/color]-bc+bc-101c+(c*c)=0Z7W;u&HOs
c*c=0y2CS[4m})| Ds
c=0V+~2R!MH,g8r!j/{]
rr S^,k ?
基於以上紅字的地方,其餘的地方我認為沒必要看了 [quote]Originally posted by [i]sakura310[/i] at 2008-1-17 09:58 PM:
-l%]Nu[h&q$h8]){
對不起
請容許我指出錯誤
在... [/quote]
係bos0|!s"}&g[+[p|
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]