Question

[Nguyễn Bảo Trân]

Cho \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)    

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

Answer 1

[Nguyễn Phạm Hồng Anh]

\(A=\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

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

Với a + b + c + d = 0      => a + b = - ( c + d )

=> \(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Với \(a+b+c+d\ne0\) => a = b = c = d

=> \(A=1+1+1+1=4\)

Answer 2

[•♡๖ۣۜLê☠๖ۣۜNɠọ¢☠๖ۣۜTυүềη☠(☠๖ۣۜTεαм☠...]

Ta có: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c}{c}=\frac{a+b+c+2d}{d}\)

\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)(1)

TH1: a + b + c + d =0

=> a + b = -c - d

     b + c = - a - d

     a + c = -b - d

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

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

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

    \(=-1+\left(-1\right)+\left(-1\right)=-3\)

TH2: \(a+b+c+d\ne0\)

Từ (1) => a = b = c =d

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

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

  \(=1+1+1=3\)