Question

cho \(\dfrac{2a+b+c+d}{a}\) =\(\dfrac{a+2b+c+d}{b}\)=\(\dfrac{a+b+2c+d}{c}\)=\(\dfrac{a+b+c+2d}{d}\)

Tính M= \(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

Answer 1

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

Từ (1) ta có: \(\Rightarrow\) \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

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

TH1: Nếu a+b+c+d #0\(\Rightarrow a=b=c=d\Rightarrow M=1+1+1+1=4\)

TH2: Nếu a+b+c+d=0 \(\Rightarrow\) a+b= -( c+d)\(\Rightarrow\) \(\dfrac{a+b}{c+d}=-1\)

Tương tự ta có:

M = -1+(-1)+(-1)+(-1)

M= -4