Giải thích các bước giải:
ĐK: $a,b,c\ne 0$
Ta có:
$a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a}(1)$
a) Khi $a=1$ thì:
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow 1 + \dfrac{1}{b} = b + \dfrac{1}{c} = c + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{b} = c\\
b + \dfrac{1}{c} = c + 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
bc = 1\\
\dfrac{{bc + 1}}{c} = c + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
bc = 1\\
\dfrac{2}{c} = c + 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
bc = 1\\
{c^2} + c - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
bc = 1\\
\left( {c - 1} \right)\left( {c + 2} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
bc = 1\\
\left[ \begin{array}{l}
c = 1\\
c = - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
c = 1;b = 1\\
c = - 2;b = \dfrac{{ - 1}}{2}
\end{array} \right.
\end{array}$
Vậy $c = 1;b = 1$ hoặc $c = - 2;b = \dfrac{{ - 1}}{2}$ khi $a=1$
b) Ta có:
$\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
a + \dfrac{1}{b} = b + \dfrac{1}{c} \Leftrightarrow a - b = \dfrac{1}{c} - \dfrac{1}{b} \Leftrightarrow a - b = \dfrac{{b - c}}{{bc}}\left( 2 \right)\\
b + \dfrac{1}{c} = c + \dfrac{1}{a} \Leftrightarrow b - c = \dfrac{1}{a} - \dfrac{1}{c} \Leftrightarrow b - c = \dfrac{{c - a}}{{ac}}\left( 3 \right)
\end{array} \right.$
$\begin{array}{l}
+ )TH1:a > b>0 \Leftrightarrow a - b > 0\\
\left( 2 \right) \Rightarrow b - c > 0 \Leftrightarrow b > c\\
\left( 3 \right) \Rightarrow c - a > 0 \Leftrightarrow c > a
\end{array}$
Như vậy $ \Rightarrow a > b > c > a\left( {mt} \right)$
$\begin{array}{l}
+ )TH2:0 < a < b \Leftrightarrow a - b < 0\\
\left( 2 \right) \Rightarrow b - c < 0 \Leftrightarrow b < c\\
\left( 3 \right) \Rightarrow c - a < 0 \Leftrightarrow c < a
\end{array}$
Như vậy $ \Rightarrow a < b < c < a\left( {mt} \right)$
Suy ra: $a=b$
Khi đó: Qua (2)$\Rightarrow b=c$
$\Rightarrow a=b=c$
Ta có đpcm.
