Áp dụng dãy tỉ số bằng nhau :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}=\dfrac{a+b+c+d}{b+c+d+a+c+d+a+b+d+b+c+a}=\dfrac{1}{3}\) \(\Rightarrow3a=b+c+d\left(1\right)\)
\(\Rightarrow3b=c+d+a\left(2\right)\)
\(\Rightarrow3c=a+b+d\left(3\right)\)
\(\Rightarrow3d=b+c+a\left(4\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow3a+3b=b+c+d+c+d+a\)
\(\Rightarrow2a+2b=2c+2d\)
\(\Rightarrow a+b=c+d\)
Từ \(\left(2\right)+\left(3\right)\Rightarrow3b+3c=a+c+d+a+b+c\)
\(\Rightarrow2b+2c=2d+2a\)
\(\Rightarrow b+c=d+a\)
Từ \(\left(1\right)+\left(3\right)\Rightarrow2a+2c=2b+2d\)
\(\Rightarrow a+c=b+d\)
Ta có :
\(b+c=a+d;a+c=b+d\)
\(\Rightarrow b+c+a+c=d+a+b+a\)
\(\Rightarrow a+b+2c=2a+a+d\)
\(\Rightarrow c=d\)
Lại có :
\(b+c=d+a;a+c=b+d\)
\(\Rightarrow b+c+b+d=d+a+a+c\)
\(\Rightarrow2b+c+d=2a+d+c\)
\(\Rightarrow a=b\)
Từ những điều trên ta thấy được :
\(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
Nguyễn Thanh Hằng Xét thiếu TH rồi bạn !!!
Ta có :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{a}{b+c+d}+1=\dfrac{b}{a+c+d}+1=\dfrac{c}{a+b+d}+1=\dfrac{d}{a+b+c}+1\\ \Rightarrow\dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{a+c+d}=\dfrac{a+b+c+d}{a+b+d}=\dfrac{a+b+c+d}{a+b+c}\)
TH1: Nếu a+b+c+d#0
thì Đỗ Thu Trà giải giống bạn Nguyễn Thanh Hằng
Nếu a+b+c+d=0 =>a+b=-(c+d); b+c=-(a+d);c+d=-(a+b); a+d=-(b+c)
Thế những cái này vao biểu thức M thì M=-4
