Trường hợp 1: a+b+c \(\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left[a+b+c\right]}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{b+c}=\frac{1}{2}\Leftrightarrow\frac{b+c}{a}=2\\\frac{b}{a+c}=\frac{1}{2}\Leftrightarrow\frac{a+c}{b}=2\\\frac{c}{a+b}=\frac{1}{2}\Leftrightarrow\frac{a+b}{c}=2\end{cases}\Rightarrow}\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2+2+2=6\)
Trường hợp 2: a + b + c = 0
\(a+b+c=0\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}}\)
\(\Rightarrow\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=-\frac{a}{a}+-\frac{b}{b}+-\frac{c}{c}=-1+\left[-1\right]+\left[-1\right]=-3\)
Ta có :
\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{a+c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2c=a+b\\2b=a+c\end{cases}\Rightarrow\hept{\begin{cases}3a=a+b+c\\3c=a+b+c\\3b=a+b+c\end{cases}\Rightarrow}a=b=c}\)
Thay a=b=c vào P :
\(P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{a+a}{a}+\frac{a+a}{a}+\frac{a+a}{a}=6\)
TH1 :\(a+b+c=0\)
thì \(\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}}\)
Suy ra : \(P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3\)
TH2: \(a+b+c\ne0\)
Thì làm như ở dưới
