\(TH1:a+b+c+d\ne0\)

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

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

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

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

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

\(=1+1+1+1\)

\(=4\)

\(TH2:a+b+c+d=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

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

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

\(=-1-1-1-1\)

\(=-4\)

Ta có:

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

⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)

    \(=\dfrac{a+b+c+2d}{d}-1\)

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

Nếu a+b+c+d=0

⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)

Thay vào M, ta có:

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

Nếu a+b+c+d ≠0

⇒ \(a=b=c=d\)

Thay vào M, ta có

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

\(\text{Cùng trừ mỗi tỉ số trên 1 đơn vị ta được:}\)

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

\(\text{Từ đây ta suy ra 2 trường hợp:}\)

\(\text{Trường hợp 1:}\)

\(\text{Nếu }a+b+c+d\notin0\Rightarrow a=b=c=d\)

\(\Rightarrow M=1+1+1+1=1.4=4\)

\(\text{Trường hợp 2:}\)

\(\text{Nếu }a+b+c+d=0\text{ thì:}\)

\(a+b=-\left(c+d\right);b+c=-\left(d+a\right)\)

\(c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)

\(\text{Do đó }M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)

hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)

\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))

\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)

\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)

\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)

\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)

\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow ac=bd\) (do \(b\ne d\))

 Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)

Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}.\dfrac{a}{b}=\dfrac{1}{2}.\dfrac{b}{c}=\dfrac{1}{2}.\dfrac{c}{d}=\dfrac{1}{2}.\dfrac{d}{a}\)

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

⇒   \(a=b=c=d\)

Thay b = a ; c = a ; d = a vào biểu thức A ta có:

\(A=\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}\)

\(A=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)

\(A=\dfrac{1}{2}.4=2\)

Vậy A = 2

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

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

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

=>\(\dfrac{a}{2b}=\dfrac{1}{2}\)=>2a=2b =>a=b

\(\dfrac{b}{2c}=\dfrac{1}{2}\)=>2b=2c =>b=c

\(\dfrac{c}{2d}=\dfrac{1}{2}\)=>2c=2d =>c=d

\(\dfrac{d}{2a}=\dfrac{1}{2}\)=>2d=2a =>d=a

=>a=b=c=d.

*\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)

=\(\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010d}{a+a}+\dfrac{2011a-2010a}{a+a}\)

=\(\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)=2

Tính cái gì vậy ạ?