想問大家一條數,,
ab+bc-ac=0a-c=101
求b}U9d.V)T4aiG
3Q&O3Z6d p Yz
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)=0
101c-bc+bc-101c+(c*c)=0)^G(g6n Yi+R
c*c=0(XbJCP(k
c=0
:k~4MX\2Ud
a=101-0#Ij:ywm9}2u
a=1014aX3kFO.X#q-l8}3t
9Vh:UXbPdX|,S$i
ab+bc-ac=0
a=1012YHI)ESl'iC0o,Ns
c=0
101b + 0*b - 101*0=0wd*k1@Z[
101b=0:id+iJoX3q6HU
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:
a=101-cq2m,Z4O!d c:l
ab+bc-ac=0 H^N_YH
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0HcW&`Kh
c*c=04q [j^NH)z+y2A
c=0
a=101-0;T3dms,r.fK9Z4O P
a=101 YUA-s6\%_ N
ab+bc-ac=0
a=101
c=0Fh V%`W sN6^
101b + 0*b - 101*0=0
101b=0
b=0 [/quote]&h`'`6Rh)M
對不起^7_7A8e'e$Sh+}I3Ev
請容許我指出錯誤
.T/T$R2eIO
在一開頭
5P#UQ!iw]1_2]+x:_
[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c)yiha)zO
ab+bc-ac=0