Đáp án:
$\begin{array}{l}
F\left( x \right) = {x^2} + 1\\
F\left( a \right) = - 2b\\
\Rightarrow {a^2} + 1 = - 2b\\
\Rightarrow {a^2} + 2b = - 1\\
F\left( b \right) = - 2c\\
\Rightarrow {b^2} + 1 = - 2c\\
\Rightarrow {b^2} + 2c = - 1\\
F\left( c \right) = - 2a\\
\Rightarrow {c^2} + 1 = - 2a\\
\Rightarrow {c^2} + 2a = - 1\\
\Rightarrow {a^2} + 2b = {b^2} + 2c = {c^2} + 2a\\
\Rightarrow {a^2} + 2b + {b^2} + 2c + {c^2} + 2a = - 1 - 1 - 1\\
\Rightarrow {a^2} + 2a + {b^2} + 2b + {c^2} + 2c = - 3\\
\Rightarrow {a^2} + 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = - 3 + 3\\
\Rightarrow {\left( {a + 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0\\
\Rightarrow {\left( {a + 1} \right)^2} = {\left( {b + 1} \right)^2} = {\left( {c + 1} \right)^2}\\
\Rightarrow a = b = c = - 1\\
A = {a^{2020}} + {b^{2020}} + {c^{2020}}\\
= {\left( { - 1} \right)^{2020}} + {\left( { - 1} \right)^{2020}} + {\left( { - 1} \right)^{2020}}\\
= 1 + 1 + 1\\
= 3
\end{array}$
Vậy A=3
