Giải thích các bước giải:
Ta có :
$\dfrac{a+b+c}{d}=\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c+b+c+d+c+d+a+d+a+b}{a+b+c+d}=3$
$\rightarrow \dfrac{a+b+c}{d}+1=\dfrac{b+c+d}{a}+1=\dfrac{c+d+a}{b}+1=\dfrac{d+a+b}{c}+1=4$
$\rightarrow \dfrac{a+b+c+d}{d}=\dfrac{a+b+c+d}{a}=\dfrac{b+c+d+a}{b}=\dfrac{c+d+a+b}{c}=4$
$\rightarrow a=b=c=d=\dfrac{1}{4}$