Question

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

CM:\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

Answer 1

Answer 2

gt: a/(b+c) + b/(c+a) + c/(a+b) = 1
A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)]
= a[a/(b+c) + 1 - 1] + b[b/(c+a) + 1 - 1] + c[c/(a+b) + 1 - 1]
= a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) - b + c.(a+b+c)/(a+b) - c
= (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] - (a+b+c)
= (a+b+c) - (a+b+c) = 0

Answer 3

a/(b-c)+b/(c-a)+c/(a-b)=0
=>a/(b-c)
=-b/(c-a)-c/(a-b)
=b/(a-b)-c/(c-a)
=(-ab+b²-c²+ac)/[(a-b)(c-a)]
=>a/(b-c)²
=(-ab+b^2-c^2+ac)/[(a-b)(b-c)(c-a)]
Tương tự:
b/(c-a)²
=(-a²+ab-bc+c²)/[(a-b)(b-c)(c-a)]
c/(a-b)²
=(-ac+a²-b²+bc)/[(a-b)(b-c)(c-a)]
Cộng các vế lại được đpcm