Question

Answer 1

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

Áp dụng t/c DTSBN:

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

\(=2\)

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

\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=3\\ \Rightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=c+d+a\\3c=d+a+b\\3d=a+b+c\end{matrix}\right.\Rightarrow a=b=c=d\)

\(\Rightarrow A=1+1^2+1^3+1^4=4\)