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

Lời giải:
Ta có:

$a(a-b)+b(b-c)+c(c-a)=a^2+b^2+c^2-ab-bc-ac$

$=\frac{3}{2}(a^2+b^2+c^2)-[\frac{1}{2}(a^2+b^2+c^2)+ab+bc+ac]$

$=\frac{3}{2}(a^2+b^2+c^2)-\frac{1}{2}(a^2+b^2+c^2+2ab+2bc+2ac)$
$=\frac{3}{2}(a^2+b^2+c^2)-\frac{1}{2}(a+b+c)^2$

$=\frac{3}{2}(a^2+b^2+c^2)$

$\Rightarrow P=\frac{a^2+b^2+c^2}{\frac{3}{2}(a^2+b^2+c^2)}=\frac{2}{3}$

P =  -a.(b-c)-b.(c-a)-c.(a-b)/(a-b).(b-c).(c-a)

   = -ab+ac-bc+ba-ca+ab/(a-b).(b-c).(c-a) = 0

Vậy P = 0 

k mk nha

Th1: a+b+c khác 0

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

\(\Rightarrow2+\frac{a+b-c}{c}=2+\frac{a-b+c}{b}=2+\frac{\left(-a\right)+b+c}{a}\)

\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)

\(\Rightarrow a=b=c\)

thay a=b=c vào b/t A. ta có:

\(A=\frac{aaa}{\left(a+a\right).\left(a+a\right).\left(a+a\right)}=\frac{aaa}{2a.2a.2a}=\frac{aaa}{8aaa}=\frac{1}{8}\)

th2: a+b+c = 0

=> a+b=-c

b+c=-a

c+a=-b

thay a+b=-c, b+c=-a, c+a=-b vào b/t A ta có:

\(A=\frac{abc}{\left(-c\right).\left(-a\right).\left(-b\right)}=-1\)