Question

已知实数a,b,c满足a/(b+c)+b/(c+a)+c/(a+b)=1,则a^2/(b+c)+b^2/(a+c)+c^

Answer 1

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

a²/(b+c)+ab/(c+a)+ac/(a+b)=a

a²/(b+c)=a-ab/(c+a)-ac/(a+b)

同理

b²/(c+a)=b-ab/(b+c)-bc/(a+b) c²/(a+b)=c-ac/(b+c)-bc/(c+a)

所以

a²/(b+c)+b²/(a+c)+c²/(a+b)

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

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

=a+b+c-a-b-c

=0