A) \[a+b+c\]
B) \[abc\]
C) 1
D) 0
Correct Answer: A
Solution :
Given, \[\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\] Let \[x=a+b+c\] \[\therefore \] \[\frac{a+b+c-a}{b+c}+\frac{a+b+c-b}{c+a}+\frac{a+b+c-c}{a+b}\] \[=\frac{b+c}{b+c}+\frac{a+c}{a+c}+\frac{a+b}{a+b}\] \[=1+1+1=3\] Hence, the value of\[x\]is \[a+b+c\].
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