$\begin{array}{l}
A = \dfrac{{2x + 1}}{{{x^2} + 2}} + \dfrac{1}{2} - \dfrac{1}{2} = \dfrac{{{x^2} + 2 + 2\left( {2x + 1} \right)}}{{2\left( {{x^2} + 2} \right)}} - \dfrac{1}{2}\\
= \dfrac{{{x^2} + 4x + 4}}{{2\left( {{x^2} + 2} \right)}} - \dfrac{1}{2} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{2\left( {{x^2} + 2} \right)}} - \dfrac{1}{2} \ge - \dfrac{1}{2}\\
\Rightarrow \min A = \dfrac{{ - 1}}{2} \Rightarrow x = - 2\\
A = \dfrac{{2x + 1}}{{{x^2} + 2}} - 1 + 1 = \dfrac{{ - {x^2} + 2x - 1}}{{{x^2} + 2}} + 1 = \dfrac{{ - {{\left( {x - 1} \right)}^2}}}{{{x^2} + 1}} + 1 \le 1\\
\Rightarrow \max A = 1 \Rightarrow x = 1\\
B = \dfrac{{3 - 4x}}{{{x^2} + 1}} + 1 - 1\\
B = \dfrac{{{x^2} - 4x + 1 + 3}}{{{x^2} + 1}} - 1 = \dfrac{{{{\left( {x - 2} \right)}^2}}}{{{x^2} + 1}} - 1 \ge - 1\\
\Rightarrow \min B = - 1 \Rightarrow x = 2\\
B = \dfrac{{3 - 4x}}{{{x^2} + 1}} - 4 + 4 = \dfrac{{ - 4{x^2} - 4x + 3 - 4}}{{{x^2} + 1}} + 4\\
= \dfrac{{ - 4{x^2} - 4x - 1}}{{{x^2} + 1}} + 4 = \dfrac{{ - {{\left( {2x + 1} \right)}^2}}}{{{x^2} + 1}} + 4 \le 4\\
\Rightarrow \max B = 4 \Rightarrow x = - \dfrac{1}{2}\\
C = \dfrac{{{x^2} + 2x + 3}}{{{x^2} + 2}} - 2 + 2\\
= \dfrac{{ - {x^2} + 2x + 3 - 4}}{{{x^2} + 2}} + 2\\
= \dfrac{{ - {x^2} + 2x - 1}}{{{x^2} + 2}} + 2 = \dfrac{{ - {{\left( {x - 1} \right)}^2}}}{{{x^2} + 2}} + 2 \le 2\\
\Rightarrow \max C = 2 \Rightarrow x = 1\\
C = \dfrac{{{x^2} + 2x + 3}}{{{x^2} + 2}} - \dfrac{1}{2} + \dfrac{1}{2}\\
C = \dfrac{{2{x^2} + 4x + 6 - {x^2} - 2}}{{2\left( {{x^2} + 2} \right)}} + \dfrac{1}{2}\\
C = \dfrac{{{x^2} + 4x + 4}}{{2\left( {{x^2} + 2} \right)}} + \dfrac{1}{2} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{2\left( {{x^2} + 2} \right)}} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\Rightarrow \min C = \dfrac{1}{2} \Rightarrow x = - 2\\
\end{array}$
