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Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\) =\(\frac{a+b+c+d}{b+c+d+a+c+d+a+b+d+a+b+c}\)
Vì a+b+c+d khác 0
=> b+c+d=a+c+d=a+b+d=a+b+c
=>a=b=c=d
Khi đó:
a + b = c+d
b+c= (a+d)
c+d=a+b
d+a=b+c
=>\(\frac{a+b}{c+d}=\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{b+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{b+a+b}{c}=\frac{a+b+c}{d}\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{b+a+b}{c}+1=\frac{a+b+c}{d}+1\)
\(=\frac{b+c+d}{a}+\frac{a}{a}=\frac{c+d+a}{b}+\frac{b}{b}=\frac{b+a+b}{c}+\frac{c}{c}=\frac{a+b+c}{d}+\frac{d}{d}\)
\(=\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
Do đó \(\frac{a+b}{c+d}+\frac{b+c}{c+d}+\frac{c+d}{a+b}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1=3\)
Ta có : \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)
Nếu a + b + c + d = 0
=> a + b = - c - d
b + c = - a - d
c + d = - b - a
d + a = - b - c
Khi đó \(P=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{d+a}+\frac{-\left(b+a\right)}{b+a}=\frac{-\left(b+c\right)}{b+c}\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Nếu a + b + c + d \(\ne\)0
\(\Rightarrow\frac{1}{c+d}=\frac{1}{d+a}=\frac{1}{b+a}=\frac{1}{b+c}\)
\(\Rightarrow c+d=d+a=b+a=b+c\)
\(\Rightarrow a=b=c=d\)
Khi đó \(P=1+1+1+1=4\)
Vậy nếu a + b + c + d = 0 thì P = - 4
nếu a + b + c + d \(\ne\)0 thì P = 4