\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

#)Giải :

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

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

TH1 : \(a+b+c=0\Leftrightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}\Leftrightarrow M=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1}\)

TH2 : \(a+b+c\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau :

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

\(\Rightarrow\hept{\begin{cases}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{cases}\Rightarrow\hept{\begin{cases}a+b=2c\\a+c=2b\\b+c=2a\end{cases}\Rightarrow}M=\frac{2c.2b.2a}{abc}=8}\)

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

\(=\frac{a+b+c}{a+b+c}=1\left(ADTCDTSBN\right)\)

\(\Rightarrow\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=2\)

\(\Rightarrow\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}=2^3=8\)

\(\Rightarrow M=8\)

Áp dụng t/c dtsbn:

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)

Ta có 

a/b+c=b/c+a=c/b+a => a/ b+c +1=b/c+a +1=c/b+a +1 

                               => a+b+c/b+c=a+b+c/c+a=a+b+c/b+a 

=> b+c=c+a=b+a 

=> a=b=c 

=> B= 2a/a+2b/b+2c/c =2+2+2=6 ( tick nhe