Question

хелп ми плиз !!!!!!!!!!!

Answer 1

$=a \left (1-\frac{1}{b+c}\right ) + b \left (1-\frac{1}{a+c}\right ) + c \left (1-\frac{1}{b+a}\right ) = a \left ( \frac {1}{a+b} + \frac{1}{a+c} \right) +b \left ( \frac {1}{a+b} + \frac{1}{b+c} \right) + c \left ( \frac {1}{b+c} + \frac{1}{a+c} \right) = \frac {a}{a+b} + \frac {a}{a+c} + \frac {b}{a+b} + \frac {b}{b+c} + \frac {c}{b+c} + \frac {c}{a+c} = \frac {a+b}{a+b} + \frac {a+c}{a+c} + \frac {b+c}{b+c} =3$