Đáp án:
$A$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
y = x\sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} \\
\Rightarrow y' = \sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} + x.\dfrac{{\left( {\dfrac{{x + 2}}{{{x^2} + 1}}} \right)'}}{{2\sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} }}\\
= \sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} + x.\dfrac{{\dfrac{{{x^2} + 1 - \left( {x + 2} \right).2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}}}{{2\sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} }}\\
= \sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} + \dfrac{{x\left( { - {x^2} - 4x + 1} \right)}}{{2{{\left( {{x^2} + 1} \right)}^2}\sqrt {\dfrac{{x + 2}}{{{x^2} + 1}}} }}\\
= \dfrac{{2{{\left( {{x^2} + 1} \right)}^2}.\dfrac{{x + 2}}{{{x^2} + 1}} + x\left( { - {x^2} - 4x + 1} \right)}}{{2\left( {{x^2} + 1} \right)\sqrt {\left( {x + 2} \right)\left( {{x^2} + 1} \right)} }}\\
= \dfrac{{2\left( {x + 2} \right)\left( {{x^2} + 1} \right) + x\left( { - {x^2} - 4x + 1} \right)}}{{2\left( {{x^2} + 1} \right)\sqrt {\left( {x + 2} \right)\left( {{x^2} + 1} \right)} }}\\
= \dfrac{{{x^3} + 3x + 4}}{{2\left( {{x^2} + 1} \right)\sqrt {{x^3} + 2{x^2} + x + 2} }}\\
\Rightarrow \left\{ \begin{array}{l}
a = 1\\
b = 2\\
c = 1\\
d = 2
\end{array} \right.\\
\Rightarrow M = a + b + c + d = 1 + 2 + 1 + 2 = 6
\end{array}$
