Ta có: \(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)

\(=\frac{a+b+c+2012d}{d}-2011\)

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

+) Xét \(a+b+c+d=0\)

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

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

\(=\frac{a+b}{-\left(a+b\right)}+\frac{b+c}{-\left(b+c\right)}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(d+a\right)}\)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

+) Xét \(a+b+c+d\) khác 0 \(\Rightarrow a=b=c=d\)

\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

Vậy...

Ta có:

\(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)\(=\frac{a+b+c+2012d}{d}\)

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

+) Xét \(a+b+c+d=0\)

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

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

\(=\frac{a+b}{-\left(a+b\right)}+\frac{b+c}{-\left(b+c\right)}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(d+a\right)}\)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

+) Xét \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

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

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

\(=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{d+a}\)

\(=1+1+1+1=4\)

Vậy ...

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Từ \(\frac{2a+b+c+d}{a}=\frac{a-2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

=> \(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

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

Nếu a+b+c+d=0 => a+b=-(c+d); b+c=-(a+d)

=> \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=-4\)

Nếu a+b+c+d \(\ne\)0 => a=c=b=d 

=> \(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}=4\)

Vì a ; b ; c ; d  > 0

=> a + b +  c + d > 0

=> 2(a + b + c + d) > 0

=> 2a + 2b + 2c + 2d > 0

Áp dụng tính chất dãy tỉ số bằng nhau

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

=> \(\frac{a}{2b}=\frac{1}{2}\Rightarrow2a=2b\Rightarrow a=b\)

Tương tự,ta được a = b = c = d

Khi đó A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)

= \(\frac{2013a-2012a}{2a}+\frac{2013b-2012b}{2b}+\frac{2013c-2012c}{2c}+\frac{2013d-2012d}{2d}\)(Vì a = b = c = d)

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

\(a,b,c,d>0\text{ nên : }a+b+c+d>0\Rightarrow\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)

do đó: a=b=c=d hay A=1/2+1/2+1/2+1/2=2

                                                       Bài giải

a, \(\left| |3x-\frac{7}{3} | -2\right|=7\)

\(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|-2=-7\\|3x-\frac{7}{3}|-2=7\end{cases}}\)                 \(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|=-5\text{ ( loại) }\\|3x-\frac{7}{3}|=9\end{cases}}\)         \(\Rightarrow\text{ }\left|3x-\frac{7}{3}\right|=9\)        \(\Rightarrow\orbr{\begin{cases}3x-\frac{7}{3}=-9\\3x-\frac{7}{3}=9\end{cases}}\)                             \(\Rightarrow\orbr{\begin{cases}3x=\frac{-20}{3}\\3x=\frac{34}{3}\end{cases}}\)                             \(\Rightarrow\orbr{\begin{cases}x=-\frac{20}{9}\\x=\frac{34}{9}\end{cases}}\)

                 \(\Rightarrow\text{ }x\in\left\{-\frac{20}{9}\text{ ; }\frac{34}{9}\right\}\)