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\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)\(\frac{a}{b+c}+1=\frac{b}{a+c}+1=\frac{c}{a+b}+1\Leftrightarrow\)\(\frac{a+b+c}{b+c}=\frac{b+a+c}{a+c}=\frac{c+a+b}{a+b}\Leftrightarrow b+c=a+c=a+b\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{a+a}{a}+\frac{a+a}{a}+\frac{a+a}{a}=2+2+2=6\)
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\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
+)Nếu a+b+c=0\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\Rightarrow B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)
Nếu \(a+b+ c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c\)
\(b+ c=2a\)
\(c+a=2b\)
\(\Rightarrow B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)
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Xét a+b+c=0 thì A=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)
Xét a+b+c\(\ne0\).Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)
\(\Rightarrow A=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a.2a.2a}{a.a.a}=8\)
Vậy.................................
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\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)
Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)
Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)
Khi đó \(P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=\dfrac{-a}{a}+\dfrac{-b}{b}+\dfrac{-c}{c}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\c+a=2b\\a+b=2c\end{matrix}\right.\)
Khi đó:
\(P=\dfrac{2a}{a}+\dfrac{2b}{b}+\dfrac{2c}{c}=2+2+2=6\)