Giải:
+) Xét a + b + c \(\ne\) 0, Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}\)
+) Xét a + b + c = 0 \(\Rightarrow-a=b+c\)
\(-b=a+c\)
\(-c=a+b\)
Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
\(\Rightarrow\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}=-1\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1\)
Vậy \(x\in\left\{\frac{1}{2};-1\right\}\)
\(x=\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
