一條maths!急!!!

1/a-1/b=1/c
1/a-1/c=1/b
c/ca-a/ac=1/b
(c-a)/ac=1/b
(c-a)b=ac
b=ac/(c-a)

1/a - 1/b = b/c
abc(1/a - 1/b) = abc(b/c)
bc - ac = ab^2
ab^2 - bc + ac = 0

b = [ -(-c)+sqrt(c^2 - 4 a^2 c) ]/2a or [ -(-c)-sqrt(c^2 - 4 a^2 c) ]/2a

b = [ c + sqrt(c^2 - 4 a^2 c) ]/2a or [ c - sqrt(c^2 - 4 a^2 c) ]/2a

2006-10-17 09:13:57 補充：
sorry 我以為你個 [b] 係 乘b的意思...it should be1/a-1/b=1/c1/a-1/c=1/b(c-a)/ac=1/bb(c-a)=acb=ac/(c-a)

2006-10-17 09:13:58 補充：
sorry 我以為你個 [b] 係 乘b的意思...it should be1/a-1/b=1/c1/a-1/c=1/b(c-a)/ac=1/bb(c-a)=acb=ac/(c-a)

1/a-1/b=1/c
→ 1/a-1/c=1/b
→ (c-a)/ac=1/b
→ ac/(c-a)=b
→ b=ac/(c-a)

唔係好明妳整個 [b] 出來做乜?
總之 b=ac/(c-a) 啦!!

1/a-1/b=1/c
(b-a)/ab=1/c
b-a=ab/c
cb-ca=ab
cb-ab=ca
b(c-a)=ca
b= ca/(c-a)
應該係咁做

1/a-1/b=1/c [b]
-1/b = 1/c + 1/a
-1/b = (a+c)/ac
之後用交义相乘
b(a+c) = -(ac)
b = -ac/(a+c)

2006-10-16 17:05:02 補充：
應係b = -ac/(a-c)

1/a-1/b=1/c
(b-a)/ab=1/c
cb-ca=ab
cb-ab=ca
b(c-a)=ca
b=ca/(c-a)
唔知岩唔岩呢???

solve 1/a-1/b=1/c

1/b = 1/a - 1/c
= (c-a)/ac
b = ac/(c-a)

其實唔肯定你個問題係咪咁..

1/a - 1/b = 1/c
1/a - 1/c = 1/b
b(1/a - 1/c) = 1
b = 1/(1/a - 1/c)
b = ac/(c-a)