Ta có \({y}'=\frac{4}{{{\left( x+2 \right)}^{2}}}\Rightarrow {y}'\left( {{x}_{0}} \right)=\frac{4}{{{\left( {{x}_{0}}+2 \right)}^{2}}}\) và \(y\left( {{x}_{0}} \right)=\frac{2{{x}_{0}}}{{{x}_{0}}+2}\) nên phương trình tiếp tuyến là

\(\left( d \right):y-y\left( {{x}_{0}} \right)={y}'\left( {{x}_{0}} \right)\left( x-{{x}_{0}} \right)\Leftrightarrow y=\frac{4}{{{\left( {{x}_{0}}+2 \right)}^{2}}}\left( x-{{x}_{0}} \right)+\frac{2{{x}_{0}}}{{{x}_{0}}+2}\Leftrightarrow \frac{4}{{{\left( {{x}_{0}}+2 \right)}^{2}}}x-y+\frac{2x_{0}^{2}}{{{\left( {{x}_{0}}+2 \right)}^{2}}}=0.\)

Khoảng cách từ điểm \(I\left( -\,2;2 \right)\) đến \(\left( d \right)\) là \(d\left( I;\left( d \right) \right)=\frac{\left| 8{{x}_{0}}+16 \right|}{\sqrt{{{\left( {{x}_{0}}+2 \right)}^{4}}+16}}=\frac{8\left| {{x}_{0}}+2 \right|}{\sqrt{{{\left( {{x}_{0}}+2 \right)}^{4}}+16}}\)

Đặt \(t=\left| {{x}_{0}}+2 \right|\,\,\left( t\ge 0 \right)\Rightarrow d\left( I;\left( d \right) \right)=\frac{8t}{\sqrt{{{t}^{4}}+16}}\)

Xét hàm số \(f\left( t \right)=\frac{t}{\sqrt{{{t}^{4}}+16}}\) trên \(\left( 0;+\infty  \right)\) ta có

\(\begin{array}{l}
f'\left( t \right) = \frac{{\sqrt {{t^4} + 16} - \frac{{t.4{t^3}}}{{2\sqrt {{t^4} + 16} }}}}{{{t^4} + 16}} = \frac{{{t^4} + 16 - 2{t^4}}}{{\left( {{t^4} + 16} \right)\sqrt {{t^4} + 16} }} = \frac{{16 - {t^4}}}{{\left( {{t^4} + 16} \right)\sqrt {{t^4} + 16} }} = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 2\,\,\left( {tm} \right)\\
t = - 2\,\,\left( {ktm} \right)
\end{array} \right.\\
f\left( 0 \right) = 0\\
f\left( 2 \right) = \frac{{\sqrt 2 }}{4}\\
\Rightarrow \mathop {\max }\limits_{\left[ {0; + \infty } \right)} f\left( t \right) = \frac{{\sqrt 2 }}{4} \Leftrightarrow f\left( t \right) \le \frac{{\sqrt 2 }}{4}\\
\Rightarrow d\left( {I;\left( d \right)} \right) \le 8.\frac{{\sqrt 2 }}{4} = 2\sqrt 2
\end{array}\)

Dấu « = » xảy ra

\(\begin{array}{l}
\Leftrightarrow t = 2 \Leftrightarrow \left| {{x_0} + 2} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
{x_0} = 0\,\,\left( {ktm} \right)\\
{x_0} = - 4\,\,\left( {tm} \right)
\end{array} \right. \Rightarrow {y_0} = \frac{{2.\left( { - 4} \right)}}{{ - 4 + 2}} = 4\\
\Rightarrow 2{x_0} + {y_0} = 2.\left( { - 4} \right) + 4 = - 4
\end{array}\)

Chọn A.