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

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

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

Vãi Phân

Đm không biết thì trả lời làm chi!!!!!!!!!!!!

Ta có kết quả tổng quát hơn như sau:

Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$

Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$