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2007-12-26 02:43:40 · 13 answers · asked by kashish 1 in Science & Mathematics Mathematics

13 answers

a,b,c =
1,1,1
2,2,2
3,3,3
...

2007-12-26 02:54:45 · answer #1 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0 1

a=b=c=-1/2

2007-12-27 03:14:54 · answer #2 · answered by Anonymous · 0 0

a=b=c not=0

2007-12-26 11:03:37 · answer #3 · answered by Arnold K 2 · 0 1

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 · answer #4 · answered by sweetu 2 · 0 0

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 · answer #5 · answered by bhatta 3 · 0 0

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 · answer #6 · answered by Baldev 5 · 2 0

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 · answer #7 · answered by Mein Hoon Na 7 · 0 0

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 · answer #8 · answered by FUNNY GUY. 4 · 1 0

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 · answer #9 · answered by bharat m 3 · 1 0

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 · answer #10 · answered by Harish 2 · 0 0

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 · answer #11 · answered by bryan 2 · 0 0

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