Ta có:
$a^2 + b^2 + c^2 + \dfrac{1}{a^2} + \dfrac{1}{b^2} +\dfrac{1}{c^2} = 6$
$\to a^2 - 2 +\dfrac{1}{a^2} + b^2 - 2 +\dfrac{1}{b^2} + c^2 -2 +\dfrac{1}{c^2} = 0$
$\to a^2 - 2a.\dfrac 1a +\dfrac{1}{a^2} + b^2 - 2b.\dfrac 1b +\dfrac{1}{b^2} + c^2 -2c.\dfrac 1c+\dfrac{1}{c^2} = 0$
$\to \left(a -\dfrac 1a\right)^2 + \left(b -\dfrac 1b\right)^2+ \left(c -\dfrac 1c\right)^2 = 0$
$\to \begin{cases}a - \dfrac 1a = 0\\b - \dfrac 1b = 0\\c - \dfrac 1c = 0\end{cases}$
$\to \begin{cases}a = \pm 1\\b = \pm 1 \\c = \pm 1\end{cases}$
Đặt $P = a^{2019} + b^{2019} + c^{2019}$
Ta có các trường hợp:
$+)\quad (a;b;c) = (1;1;1)$
$\to P = 1 + 1 + 1 = 3$
$+) \quad (a;b;c) = (-1;-1;-1)$
$\to P = - 1 - 1 - 1 = - 3$
$+) \quad (a;b;c) = (-1;1;1),(1;-1;1),(1;1-1)$
$\to P = -1 + 1 +1 = 1- 1 + 1 = 1 + 1 - 1= 1$
$+) \quad (a;b;c)=(-1;-1;1),(-1;1;-1),(1;-1;-1)$
$\to P = -1 - 1 + 1 = -1+ 1- 1 = 1 - 1 - 1= -1$
