Đáp án:
$abc=1$ hoặc $abc=-1$
Giải thích các bước giải:
ĐK: $a\ne b\ne c;1+ab\ne 0; 1+bc\ne 0; 1+ca\ne 0$
Ta có:
$\dfrac{a}{{1 + ab}} = \dfrac{b}{{1 + bc}} = \dfrac{c}{{1 + ca}}(1)$
+)TH1: $a= 0$
$ \Rightarrow \dfrac{b}{{1 + bc}} = \dfrac{c}{{1 + ca}} = 0 \Rightarrow b = c = 0\left( mt \right)$
+)TH2: $a\ne 0$
Khi đó:
$ \dfrac{b}{{1 + bc}} = \dfrac{c}{{1 + ca}} \ne 0 \Rightarrow b, c \ne 0$
$\begin{array}{l}
\dfrac{a}{{1 + ab}} = \dfrac{b}{{1 + bc}} = \dfrac{c}{{1 + ca}}\\
\Rightarrow \dfrac{{1 + ab}}{a} = \dfrac{{1 + bc}}{b} = \dfrac{{1 + ca}}{c}\\
\Rightarrow b + \dfrac{1}{a} = c + \dfrac{1}{b} = a + \dfrac{1}{c}\\
\Rightarrow \left\{ \begin{array}{l}
b - c = \dfrac{1}{b} - \dfrac{1}{a} = \dfrac{{a - b}}{{ab}}\\
c - a = \dfrac{1}{c} - \dfrac{1}{b} = \dfrac{{b - c}}{{bc}}\\
b - a = \dfrac{1}{c} - \dfrac{1}{a} = \dfrac{{a - c}}{{ac}}
\end{array} \right.\\
\Rightarrow \left( {b - c} \right)\left( {c - a} \right)\left( {b - a} \right) = \dfrac{{a - b}}{{ab}}.\dfrac{{b - c}}{{bc}}.\dfrac{{a - c}}{{ac}}\\
\Rightarrow \left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right) = \dfrac{{\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}{{{{\left( {abc} \right)}^2}}}\\
\Rightarrow \dfrac{1}{{{{\left( {abc} \right)}^2}}} = 1\left( {Do:a \ne b \ne c \Rightarrow \left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right) \ne 0} \right)\\
\Rightarrow {\left( {abc} \right)^2} = 1\\
\Rightarrow \left[ \begin{array}{l}
abc = 1\\
abc = - 1
\end{array} \right.
\end{array}$
Như vậy: $abc=1$ hoặc $abc=-1$
Vậy $abc=1$ hoặc $abc=-1$
