想問大家一條數,,
ab+bc-ac=0a-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-c6Av6s:rD6MZY
ab+bc-ac=0+EhX@`.CC$m[:AZ:W:e
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0
c*c=0t5M/q:i,YsA$|D
c=0u%q2` a @o(Q
Y(OA m;}lEQg
a=101-0U2M-[mU,L7ixr
a=101bQ6aR"ER J
ab+bc-ac=0
a=101
c=0
101b + 0*b - 101*0=0
101b=0
b=0 [quote]Originally posted by [i]ch.chak[/i] at 2008-1-17 09:38 PM:
a=101-c:Pr(lU*\G(Rr
ab+bc-ac=0
b(101-c)+bc-c(101-c)=0
101c-bc+bc-101c+(c*c)=0
c*c=0{-}FyL$kKpgZ
c=0w:Ru5D;_P'ZlL7L
a=101-0^L(hd^o5_6d^
a=101,aI*sA+D:t1u
ab+bc-ac=01x$lE Y)F)^
a=101zG:q-R3h|0Y2h'A ~*n
c=0;J;YF M;UU
101b + 0*b - 101*0=03V9S4M)?'Sq0t!{
101b=0
b=0 [/quote]/E4iU q} G3b^Cd"o
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在一開頭
[color=Red]a=101-c[/color](原本是a-c=101,移項應該是a=101+c),N ?{#yB5E`1wP;X,f
ab+bc-ac=0
[color=Red]b[/color]([color=Red]101[/color]-c)+bc-c(101-c)=0