Đáp án:
$P=81$
Giải thích các bước giải:
$\dfrac{a+b+c-d}{d}=\dfrac{b+c+d-a}{a}=\dfrac{c+d+a-b}{b}=\dfrac{d+a+b-c}{c}=\dfrac{(a+b+c-d)+(b+c+d-a)+(c+d+a-b)+(d+a+b-c)}{d+a+b+c}=\dfrac{2(a+b+c+d)}{a+b+c+d}=2$
$\rightarrow \dfrac{a+b+c}{d}-1=\dfrac{b+c+d}{a}-1=\dfrac{c+d+a}{b}-1=\dfrac{d+a+b}{c}-1=2$
$\rightarrow \dfrac{a+b+c}{d}=\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=3 $
$\rightarrow \dfrac{a+b+c}{d}.\dfrac{b+c+d}{a}.\dfrac{c+d+a}{b}.\dfrac{d+a+b}{c}=3^4$
$\rightarrow \dfrac{a+b+c}{a}.\dfrac{b+c+d}{b}.\dfrac{c+d+a}{c}.\dfrac{d+a+b}{d}=3^4$
$\rightarrow(1+ \dfrac{b+c}{a}).(1+\dfrac{c+d}{b}).(1+\dfrac{d+a}{c}).(1+\dfrac{a+b}{d})=3^4$
$\rightarrow P=81$