Question

Cho a+b+c+d\(\ne\)0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)

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

Answer 1

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}\dfrac{1}{3}\)(vìa+b+c+d\(\ne\)0)

=>3a=b+c+d: 3b=a+c+d=>3a-3b=b-a

=>3(a-b)=-(a-b)=>4(a-b)=0=>a=b

Tương tự => a=b=c=d=> A=4

Answer 2

Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)

Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{a+b}{a+b+2\left(c+d\right)}=\dfrac{1}{3}\)

\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)

\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\)

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

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

Tương tự:\(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)

Vậy A=4.