Với \(a,b\) là các hằng số, hệ số tự do của đa thức \({x^3} - 7\left( {a + 1} \right){x^2} - {a^2} + {b^2} - ab + 3\) là:

\(f\left( 1 \right) = 1011;f\left( { - 1} \right) =  - 1011\)

\(f\left( 1 \right) = 1011;f\left( { - 1} \right) = 1011\)

\(f\left( 1 \right) = 1011;f\left( { - 1} \right) =  - 1009\)

\(f\left( 1 \right) = 2021;f\left( { - 1} \right) = 2021\)

\(14\)

\(9\)

\(5\)

\(7\)

\(f\left( x \right) = 3x + 4.\)

\(f\left( x \right) = \dfrac{8}{3}x - 4.\)

\(f\left( x \right) = \dfrac{8}{3}x + 4.\)

\(f\left( x \right) =  - \dfrac{8}{3}x + 12.\)

\(f\left( x \right) = x + 2\)

\(f\left( x \right) =  - x + 4\)

\(f\left( x \right) =  - \dfrac{7}{3}x + \dfrac{{16}}{3}\)

\(f\left( x \right) = x - 2\)

\(a =  - \dfrac{2}{5}\)

\(a = \dfrac{5}{2}\)

\(a = 4\)

\(a = \dfrac{2}{5}\)

\(f\left( { - 2} \right) = g\left( { - 2} \right)\)

\(f\left( { - 2} \right) = 3.g\left( { - 2} \right)\)

\(f\left( { - 2} \right) > g\left( { - 2} \right)\)

\(f\left( { - 2} \right) < g\left( { - 2} \right)\)

\(f\left( 0 \right) = g\left( 1 \right)\)

\(f\left( 0 \right) > g\left( 1 \right)\)

\(f\left( 0 \right) < g\left( 1 \right)\)

\(f\left( 0 \right) \ge g\left( 1 \right)\)