a) $TXĐ: D = \Bbb R$
b) $f(0) = \dfrac{2.0 + 1}{0 + 2} = \dfrac{1}{2}$
$f(2) = \dfrac{2.2 + 1}{2 + 2} = \dfrac{5}{4}$
$f(-1) = \dfrac{\sqrt[3]{2.(-1) + 1}}{-1 - 1} = \dfrac{1}{2}$
$f(-3) = \dfrac{\sqrt[3]{2.(-3) + 1}}{-3 - 1} =\dfrac{\sqrt[3]{5}}{4}$