Question

if ....., then find the value of x

Answer 1

HEY DEAR YOUR ANSWER IS HERE
C. -(a+b+c) is the right option...

Answer 2

Answer:

-(a+b+c)

Step-by-step explanation:

Given : $\frac{x+a}{b+c}+\frac{x+b}{a+c}+\frac{x+c}{a+b} + 3 = 0$

To Find : x

Solution :

$\frac{x+a}{b+c}+\frac{x+b}{a+c}+\frac{x+c}{a+b} + 3 = 0$

$\frac{x+a}{b+c}+1+\frac{x+b}{a+c}+1+\frac{x+c}{a+b} + 1 = 0$

$\frac{x+a+b+c}{b+c}+\frac{x+b+a+c}{a+c}+\frac{x+c+a+b}{a+b}= 0$

$(x+a+b+c)(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})= 0$

$(x+a+b+c)= 0$

$x=-(a+b+c)$

Hence The value of x is -(a+b+c)