\(\dfrac{a}{b+c+d}=\dfrac{b}{c+a+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
+)Xét a+b+c+d=0 thì a+d=-c-d
b+c=-d-a
c+d=-b-a
d+a=-b-c
Do đó:
\(P=\dfrac{-c-d}{c+d}+\dfrac{-a-b}{a+b}+\dfrac{-b-c}{b+c}+\dfrac{-d-a}{a+d}\\ =-1+-1+-1+-1=-4\)
+)Xét a+b+c+d khác 0
áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+c+d}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
=>\(a=\dfrac{1}{3}\left(d+b+c\right)\)
\(b=\dfrac{1}{3}\left(a+c+d\right)\)
\(c=\dfrac{1}{3}\left(a+b+d\right)\)
\(d=\dfrac{1}{3}\left(a+b+c\right)\)
Bạn thay vào r tính
Ta có : \(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}\)
\(\Rightarrow\)\(\dfrac{a}{b+c+d}+1=\dfrac{b}{c+d+a}+1=\dfrac{c}{d+a+b}+1=\dfrac{d}{a+b+c}+1\)
\(\Rightarrow\)\(\dfrac{a+b+c+d}{b+c+d}=\dfrac{b+c+d+a}{c+d+a}=\dfrac{c+d+a+b}{d+a+b}=\dfrac{d+a+b+c}{a+b+c}\)
TH1 : \(a+b+c+d\ne0\)\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) P= \(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
TH2 : \(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)