Đáp án:
$A =\dfrac{1}{2017^{2017}}$
Giải thích các bước giải:
Sửa đề:
$\quad \begin{cases}ab + bc + ca = 2017abc\\2017(a+b+c) = 1\end{cases}$
$\to \begin{cases}\dfrac{ab + bc + ca}{abc}=2017\\\dfrac{1}{a+b+c}=2017\end{cases}$
$\to \dfrac1a +\dfrac1b +\dfrac1c =\dfrac{1}{a+b+c}$
$\to \dfrac{a + b}{ab} +\dfrac1c -\dfrac{1}{a+b+c}=0$
$\to \dfrac{a + b}{ab} +\dfrac{a + b}{c(a + b + c)}= 0$
$\to (a+b)\cdot\left(\dfrac{1}{ab} +\dfrac{1}{c(a+b+c)}\right)=0$
$\to (a+b)\cdot\dfrac{c(a + b + c) + ab}{abc(a+b+c)}=0$
$\to (a+b)\cdot\dfrac{(ac + ab)+ c(b+c)}{abc(a+b+c)}=0$
$\to (a+b)(b+c)(c+a) = 0$
$\to \left[\begin{array}{l}\begin{cases}a = - b\\c = \dfrac{1}{2017}\end{cases}\\\begin{cases}b = - c\\a = \dfrac{1}{2017}\end{cases}\\\begin{cases}c= - a\\b= \dfrac{1}{2017}\end{cases}\end{array}\right.$
$\to A = \left[\begin{array}{l}c^{2017} =\dfrac{1}{2017^{2017}}\\a^{2017} =\dfrac{1}{2017^{2017}}\\b^{2017} =\dfrac{1}{2017^{2017}}\end{array}\right.$
$\to A =\dfrac{1}{2017^{2017}}$
