Đáp án:
d. \(\left[ \begin{array}{l}
D = 1\\
D = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
c.C = 4\sqrt x - \dfrac{{{{\left( {\sqrt x + 3} \right)}^2}\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\sqrt {{{\left( {x - 2} \right)}^2}} \\
= 4\sqrt x - \left( {\sqrt x + 3} \right).\left| {x - 2} \right|\\
\to \left[ \begin{array}{l}
C = 4\sqrt x - \left( {\sqrt x + 3} \right).\left( {x - 2} \right)\left( {DK:2 \le x \ne 9} \right)\\
C = 4\sqrt x - \left( {\sqrt x + 3} \right).\left( { - x + 2} \right)\left( {DK:0 \le x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = 4\sqrt x - x\sqrt x - 3x + 2\sqrt x + 6\\
C = 4\sqrt x + x\sqrt x + 3x - 2\sqrt x - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = - x\sqrt x - 3x + 6\sqrt x + 6\\
C = x\sqrt x + 3x + 2\sqrt x - 6
\end{array} \right.\\
d.D = \dfrac{{\sqrt {{{\left( {x + 2} \right)}^2}} }}{{\left( {x + 2} \right)}} = \dfrac{{\left| {x + 2} \right|}}{{x + 2}}\\
\to \left[ \begin{array}{l}
D = \dfrac{{x + 2}}{{x + 2}}\left( {DK:x > - 2} \right)\\
D = - \dfrac{{x + 2}}{{x + 2}}\left( {DK:x < - 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
D = 1\\
D = - 1
\end{array} \right.
\end{array}\)
