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Áp dụng t/c dtsbn ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2b+2c+2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)
\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)
\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)
\(TH1:a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=1+1+1+1\)
\(=4\)
\(TH2:a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)
\(=-1-1-1-1\)
\(=-4\)
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)
\(=\dfrac{a+b+c+2d}{d}-1\)
⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Nếu a+b+c+d=0
⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)
Thay vào M, ta có:
\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)
Nếu a+b+c+d ≠0
⇒ \(a=b=c=d\)
Thay vào M, ta có
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
Vì \(a,b,c>0\Rightarrow a+b+c\ne0\)
Áp dụng tc dtsbn:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Rightarrow P=\dfrac{abc}{2a\cdot2b\cdot2c}=\dfrac{1}{8}\)
TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(P=\dfrac{\left(b+c\right)}{b}.\dfrac{\left(a+b\right)}{a}.\dfrac{\left(a+c\right)}{c}=\dfrac{-a}{b}.\dfrac{-c}{a}.\dfrac{-b}{c}=-1\)
TH2: \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a-b+c}{2b}=\dfrac{c-a+b}{2a}=\dfrac{a-c+b}{2c}=\dfrac{a-b+c+c-a+b+a-c+b}{2b+2a+2c}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b+c}{2b}=\dfrac{1}{2}\\\dfrac{c-a+b}{2a}=\dfrac{1}{2}\\\dfrac{a-c+b}{2c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+c=2b\\c+b=2a\\a+b=2c\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)