Question

if a2= b+c, b2= c+a, c2= a+b, then find the value of 1/1+a+1/1+b+1/1+c

Answer 1

1/1+a+1/1+b+1/1+c
=a/a+a²+b/b+b²+c/c+c²
=a/a+b+c + b/a+b+c + c/a+b+c
=a+b+c/a+b+c
=1
value is 1
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Answer 2

Thus, The value of  $\frac{1}{1+a} +\frac{1}{1+b} +\frac{1}{1+c}$  is $1$

Step-by-step explanation:

Given;

$a^{2}=b+c$ , $b^{2}=c+a$ , $c^{2} =a+b$ then $\frac{1}{1+a} +\frac{1}{1+b} +\frac{1}{1+c} =?$

∴$\frac{1}{1+a} +\frac{1}{1+b} +\frac{1}{1+c}$

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

$=\frac{a}{a+a^{2} } +\frac{b}{b+b^{2} } +\frac{c}{c+c^{2} }$

Plug $a^{2}$ , $b^{2}$ and $c^{2}$ value in above equation;

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

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

$=1$

∴ The value of  $\frac{1}{1+a} +\frac{1}{1+b} +\frac{1}{1+c}$  is $1$