ta có \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
=> \(\left(\dfrac{a}{b+c+d}+1\right)=\left(\dfrac{b}{a+c+d}+1\right)=\left(\dfrac{c}{a+b+d}+1\right)=\left(\dfrac{d}{a+b+c}+1\right)\)
(=) \(\dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{a+c+d}=\dfrac{a+b+c+d}{a+b+d}=\dfrac{a+b+c+d}{a+b+c}\)
*Nếu a+b+c+d=0
=> \(\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{matrix}\right.\)
=> M=(-1)+(-1)+(-1)+(-1)=(-4)
Nếu a+b+c+d\(\ne\)0
=> a=b=c=d
=> M=1+1+1+1=4
Xét a+b+c+d=0
\(\Rightarrow\)a=-(b+c+d).Thay vào \(\dfrac{a}{b+c+d}\)ta có
\(\dfrac{-\left(b+c+d\right)}{b+c+d}\)=-1.Làm tương tự như thế ta có
M=-1+(-1)+(-1)+(-1)=-4
Xét a+b+c+d\(\ne\)0
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b+c+d}\)=\(\dfrac{b}{a+c+d}\)=\(\dfrac{c}{a+b+d}\)=\(\dfrac{d}{b+c+a}\)
=\(\dfrac{a+b+c+d}{2\cdot\left(a+b+c+d\right)}\)=\(\dfrac{1}{2}\)
Vì\(\dfrac{a}{b+c+d}\)=\(\dfrac{1}{2}\)
\(\Rightarrow\)2a=b+c+d
\(\Rightarrow\)3a=a+b+c+d\(\left(1\right)\)
Vì\(\dfrac{b}{a+c+d}\)=\(\dfrac{1}{2}\)
\(\Rightarrow\)2b= a+c+d
\(\Rightarrow\)3b=a+b+c+d\(\left(2\right)\)
Vì\(\dfrac{c}{a+b+d}\)=\(\dfrac{1}{2}\)
\(\Rightarrow\)2c=a+b+d
\(\Rightarrow\)3c=a+b+c+d\(\left(3\right)\)
Vì\(\dfrac{d}{b+c+a}\)=\(\dfrac{1}{2}\)
\(\Rightarrow\)2d=b+c+a
\(\Rightarrow\)3d=a+b+c+d\(\left(4\right)\)
Từ\(\left(1\right)\),\(\left(2\right)\),\(\left(3\right)\),\(\left(4\right)\)
\(\Rightarrow\)3a=3b=3c=3d
\(\Rightarrow\)a=b=c=d.Khi đó
M=\(\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}\)
=1+1+1+1
=4
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