Question

Answer 1

Lời giải:

Đặt \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=t\) với $t>0$

Ta có: \(t^4=\frac{a}{2b}.\frac{b}{2c}.\frac{c}{2d}.\frac{d}{2a}=\frac{1}{16}\) \(\Rightarrow t=\frac{1}{2}\) (do $t>0$)

$\Rightarrow \frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{1}{2}$

$\Rightarrow a=b=c=d$

Khi đó:

\(A=\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}\)

\(=\frac{a}{2a}.4=2\)