a,b,c =
1,1,1
2,2,2
3,3,3
...
2007-12-26 02:54:45
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answer #1
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answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6
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a=b=c=-1/2
2007-12-27 03:14:54
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answer #2
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answered by Anonymous
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a=b=c not=0
2007-12-26 11:03:37
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answer #3
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answered by Arnold K 2
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If a=0 & b=0 & c=0
then a/b+c=b/a+c=c/a+b
n other
a=1 & b=1 & c=1
then also a/b+c=b/a+c=c/a+b
So if a& B & C = same value then ur puzzle solve
okkkkkkkkkkkkkk
Sweeetuuu
2007-12-26 23:58:59
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answer #4
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answered by sweetu 2
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Observe these ratios. you notice that these are cyclic i.e, if you replace a with b..b with c, and c with a , you get the successive ratios.
therefore a=b=c is the solution. Each can be any number. but all the three equal.
Obviously you have to eliminate the solution 0 since then the ratio becames indeterminate.
Therefore a=b=c, They can be any number except 0 .
This is the solution.
2007-12-28 03:39:47
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answer #5
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answered by bhatta 3
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Hi, it seems like you are plaining to sit in some enterance exam or what. well here's the sollution,
taking the first two equations
a/b +c = b/a +c
=> (a+bc)/b = (b+ca)/a
=> a*a +abc = b*b +abc (now abc will canncel each other)
=> a*a = b*b
=> (a=b)
now taking the first and the last equation
a/b + c = c/a + b
=> (a+bc)/b = (c+ba)/a (using the first result a=b)
=> (b+bc)/b = (c+bb)/b (replacing a by b)
=> b+bc = c+bb (rearranging)
=> b-bb = c-bc
=> b(1-b) = c(1-b)
=> b=c
So the sollution is a=b=c
2007-12-26 23:21:32
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answer #6
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answered by Baldev 5
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take reciprocal as a b and c non zero
(b+c)/a = (a+c)/b = (a+b)/c
add 1
(b+c+a)/a = (b+c+a)/b = (b+c+a)/c
so a+b+c = 0 or 1/a = 1/b = 1/c => a= b= c
2007-12-29 21:48:09
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answer #7
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answered by Mein Hoon Na 7
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Let each ratio is equal to k
=> a = k(b + c)
b = k(a + c)
& c = k(a + b)
adding we get
a+b+c = 2k(a+b+c)
=> k = 1/2
so each ratio is equal to 1/2
=> 2a = b+c,2b= a+c,2c= a+b
THIS TYPE HAS NO PARTICULAR ANSWER !!
2007-12-26 11:39:01
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answer #8
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answered by FUNNY GUY. 4
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a/(b+c) =b/(a+c) =c/(a+b)
Let each ratio is equal to k
=> a = k(b + c)
b = k(a + c)
& c = k(a + b)
adding we get
a+b+c = 2k(a+b+c)
=> k = 1/2
so each ratio is equal to 1/2
=> 2a = b+c,2b= a+c,2c= a+b
but these equations do not have a unique solution
2007-12-26 11:07:23
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answer #9
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answered by bharat m 3
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i hope my answer is right if not my apologies
a=b=c=1;
try substituting the values u will know it.
2007-12-26 11:37:46
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answer #10
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answered by Harish 2
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from: a/b+c=b/a+c > a/b=b/a > a^2=b^2 > a=b
from: b/a+c=c/a+b > (b+ac)/a=(c+ab)/a > b+ac=c+ab
subst a=b into b+ac=c+ab: a+ac=c+a^2 > c=(a^2-a)/(a-1) > c=a(a-1)/(a-1) > c=a
Therefore, a=b=c
2007-12-26 11:13:08
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answer #11
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answered by bryan 2
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