Question

[HanSoo >>>^^^.^^^<<<]

Cho a, b, c, d thỏa:

\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\)\(\frac{d}{a+b+c}\)

Tính P = \(\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{b+a}=\)\(\frac{d+a}{b+c}\)

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Answer 1

[Xyz OLM]

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