Tính (a+b)/(c+d)+(b+c)/(d+a)+(c+d)/(a+b)+(d+a)/(b+c) biết (2a+b+c+d)/a

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

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 \(A=\left[{}\begin{matrix}4\\-4\end{matrix}\right.\)