Question

Cho \(a+b+c+d\ne0\)và \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

Tìm giá trị của A = \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)

Answer 1

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

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

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

Mà a + b + c + d \(\ne\)0  \(\Rightarrow\) \(b+c+d=a+c+d=a+b+d=a+b+c\)

                                       \(\Rightarrow\)     \(a=b=c=d\)

                                      \(\Rightarrow\) \(M=4\)