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TH1: \(a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow a=b=c=d\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=1+1+1+1\)
\(\Rightarrow P=4\)
TH2: \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-4\)
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\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu \(a+b+c+d\ne0\Rightarrow a=b=c=d\)
\(\Rightarrow M=1+1+1+1=4\)
Nếu a + b + c + d = 0 => a + b = -(c + d) ; (b + c) = -(a + d) ; c + d = -(a+b) ; d + a = -(b + c)
\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Vậy M = 4 hoặc M = -4
ta có;\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)\(=>\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)\(=>\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)=>\(\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\)
Nếu a=b=c=d=0=>\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+a}{a+b}+\dfrac{a+a}{b+c}=0\)
Nếu a,b,c,d≠0=>4a=4b=4c=4d
=>a=b=c=d
do đó;\(\dfrac{a+a}{a+a}+\dfrac{b+b}{b+b}+\dfrac{c+c}{c+c}+\dfrac{d+d}{d+d}=1+1+1+1=4\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Ta sẽ chứng minh phương trình sau chỉ đúng khi: \(\left[{}\begin{matrix}a+b+c+d=0\\a=b=c=d\end{matrix}\right.\)
Thật vậy:
Từ: \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}\Leftrightarrow a\left(a+b+c+d\right)=b\left(a+b+c+d\right)\)\(\Leftrightarrow\left(a-b\right)\left(a+b+c+d\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b+c+d=0\end{matrix}\right.\)(1)
Từ: \(\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}\Leftrightarrow b\left(a+b+c+d\right)=c\left(a+b+c+d\right)\Leftrightarrow\left(b-c\right)\left(a+b+c+d\right)\Leftrightarrow\left[{}\begin{matrix}b=c\\a+b+c+d=0\end{matrix}\right.\)(2)
Từ: \(\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\Leftrightarrow c\left(a+b+c+d\right)=d\left(a+b+c+d\right)\Leftrightarrow\left(c-d\right)\left(a+b+c+d\right)=0\Leftrightarrow\left[{}\begin{matrix}c=d\\a+b+c+d=0\end{matrix}\right.\)(3)
Phương trình cần chứng minh cần thỏa mãn cả 3 phương trình (1);(2);(3),hay \(\left[{}\begin{matrix}a+b+c+d=0\\a=b=c=d\end{matrix}\right.\)
\(\circledast\) Với \(a+b+c+d=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\Leftrightarrow A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
\(\circledast\) Với \(a=b=c=d\Leftrightarrow P=1+1+1+1=4\)
Vậy \(A=\left[{}\begin{matrix}4\\-4\end{matrix}\right.\)
Theo đề bài, ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\) vì a,b,c,d khác 0
\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=1+1+1+1=4\)