想問大家一條數,,
ab+bc-ac=0RiY[f-F4^A%fW'pa-c=101
求b
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要STEPS,,,THX a=101, b&c =0 :D 我都知,,但唔知點得出個ans.,,anyway,,THX a=101-cV+j3S.d+h/s
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0^o8Fh1h!~)uS
c*c=0OPe$u x
c=0
a=101-0,P2S#_ Y;IR![q@E(O
a=101
m3Z z"c5G
ab+bc-ac=0-@^ uX.V,@
a=1012J`x#i k)_*^U8Qzo3u
c=0)b8J)@6G9P-boe8o
101b + 0*b - 101*0=0
101b=0i%NEKB:b4L
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:(oC[fW$s^4W1R1\8_
a=101-c
ab+bc-ac=0l l$]6YN5~ E
b(101-c)+bc-c(101-c)=0ln ioGr
101c-bc+bc-101c+(c*c)=0
c*c=0
c=0
"b _ x5_(H*an5|
a=101-0
a=101CPm"vc
ab+bc-ac=0l!{Pc2_4g@e
a=101e7\*d.b!k:B2[w4u+h
c=0
101b + 0*b - 101*0=0Xki6^1sl@
101b=0!Wm` \h7O j*B ~n
b=0 [/quote]
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對不起
請容許我指出錯誤V-Y5uN7Q
v%Z;Br AYY
在一開頭5qbx&M@c,A
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[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)9? Vd-uX,s;~rH
ab+bc-ac=05Ln"V;U8_ e
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0,N7Ic W1{+`
[color=Red]101c[/color]-bc+bc-101c+(c*c)=00H*^KO%eV:b
c*c=09E2l9e I2o3V1CE
c=00a"f5~U9pvNd
基於以上紅字的地方,其餘的地方我認為沒必要看了 [quote]Originally posted by [i]sakura310[/i] at 2008-1-17 09:58 PM:yC0W6y |8To
3|#K{7J&W!e"@0@
對不起(JV gO_x.UI
請容許我指出錯誤
x],}*[3fB9oU]V
在... [/quote]x'Lt/IfUZ%f
係bo
冇留意錯左tim [quote]Originally posted by [i]托迪仔T[/i] at 2008-1-17 07:42 PM:
a=101, b&c =0 :D [/quote]
ab+bc-ac=07G1mB8hWE:J+h
a-c=101#L h4[N O*vA4U Z)z
求b
咁玩法...如果c係-101...a同b=0 都得架wo:D [quote]Originally posted by [i]sakura310[/i] at 09:58 PM:$j"d!LB&e6HwY~5_"M
對不起
請容許我指出錯誤d Xx5P.TB6p8g FE
7f5C7Ii)H,h
在... [/quote]