Question

чему равно a^2/(b+c)+b^2/(a+c)+c^2/(a+b) /(a+b+c) если a/(b+c)+b/(a+c)+c/(a+b)=0

Answer 1

a/(b+c) + b/(a+c) + c/(a+b) = 0 ⇒

a/(b+c) = -(b/(a+c) + c/(a+b))

b/(a+c) = -(a/(b+c) + c/(a+b))

c/(a+b) = -(a/(b+c) + b/(a+c))

Тогда (a²/(b+c) + b²/(a+c) + c²/(a+b))/(a+b+c) =

(-a·(b/(a+c) + c/(a+b)) - b·(a/(b+c) + c/(a+b)) - c·(a/(b+c) + b/(a+c)))/(a+b+c) =

(-ab/(a+c) - ac/(a+b) - ab/(b+c) - bc/(a+b) - ac/(b+c) - bc/(a+c))/(a+b+c) =

(-(ab+bc)/(a+c) - (ac+bc)/(a+b) - (ab+ac)/(b+c))/(a+b+c) =

(-b·(a+c)/(a+c) - c·(a+b)/(a+b) - a·(b+c)/(b+c))/(a+b+c) = (-b-c-a)/(a+b+c) =

-(b+c+a)/(a+b+c) = -1