Question

If $\frac{x+a}{b+c}+\frac{x+b}{c+a}+\frac{x+c}{a+b}+3=0,a>0,b>0,c>0$, then the value of x is:

• A. $-(a^2+b^2+c^2)$
• B. $(a+b+c)$
• C. $-(a+b+c)$
• D. $\sqrt{a+b+c}$

Answer 1

Answer: (C)

$-(a+b+c)$

$\frac{x+a}{b+c}+1+\frac{x+b}{c+a}+1+\frac{x+c}{a+b}+1=0$
$\Rightarrow \frac{x+a+b+c}{b+c}+\frac{x+a+b+c}{c+a}+\frac{x+a+b+c}{a+b}=0$
$\Rightarrow (x+a+b+c)[\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}]=0$
$\Rightarrow x+a+b+c=0$
$\Rightarrow x=-(a+b+c)$