Interesting triples!Hmmmm!

Level 2

If a a , b b and c c are real numbers such that :

a − c b + c + b − a c + a + c − b a + b = 1 \dfrac{a-c}{b+c}+\dfrac{b-a}{c+a}+\dfrac{c-b}{a+b}=1 then what is the value of : a + b b + c + b + c c + a + c + a a + b = \dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}= ?


The answer is 4.

Lorenc Bushi by Lorenc Bushi , 21, Albania

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Vijay Katta
Jan 14, 2014

add 1 to each term on left side

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...